A doctor orders Quinidine for an adult patient weighing 110 lb at a dosage of 25 mg/kg/day q6h. How many
milligrams should the patient receive each day?​

Answers

Answer 1

Answer:

Total amount receive each day = 1250 mg per day

Number of dosage = 1250 / 4 = 312.5 mg per meal

Step-by-step explanation:

Given:

Weight of patient = 110 lb

Dosage = 25 mg/kg/day

Find:

Total amount receive each day

Computation:

Weight of patient = 110 lb

1 lb = 0.453592

Weight of patient = 110 (0.453592)

Weight of patient = 49.89

Weight of patient = 50 kg (Approx)

Total amount receive each day = 50 kg × 25 mg/kg/day

Total amount receive each day = 1250 mg per day

Number of dosage = 1250 / 4 = 312.5 mg per meal


Related Questions

Which equation does the graph of the systems of equations solve? (1 point) 2 linear graphs. They intersect at negative 1, 1

Answers

Answer:

  3x +4 = -2x -1

Step-by-step explanation:

The line that goes up to the right has a y-intercept of +4. This is where it crosses the y-axis. It's slope (rise/run) is 3/1 = 3, so its equation in slope-intercept form is ...

  y = mx +b . . . . where m is the slope, b is the y-intercept

  y = 3x +4

The other line has a negative slope and a y-intercept of -1. The slope of that line is rise/run = -2/1 = -2, so its equation is ...

  y = -2x -1

__

The solution point will have the x-coordinate that is the solution of the equation ...

  y = y

  3x +4 = -2x -1 . . . . . . substituting the above expressions for y.

Simplify your answer as much as possible

Answers

You said    - 1/3 - 3/5 x  =  1/2

Multiply each side by 3 :

- 1 - 9/5 x  =  3/2

Multiply each side by 5 :

- 5 - 9x  =  15/2

Multiply each side by 2 :

- 10 - 18x = 15

Add 10 to each side :

- 18x  =  25

Divide each side by -18 :

x = - 25/18

or  x = - 1 and 7/18 (same thing)

The coffee cups can hold 7/9 of a pint of liquid. If Emily pours 2/3 of a pint of coffee into a cup,how much milk can a customer add? PLZ HELP!​

Answers

Answer:

1/9

Step-by-step explanation:

easy 2/3 is equivalent to 6/9. So there is 1/9 of a pint left

Answer this will give 10 points

Answers

Answer:

maximum --> 62

median --> 46.5

upper quartile --> 60

lower quartile --> 37

minimum --> 32

Step-by-step explanation:

Forgive me on the explanation as I'm a bit rusty on these types of problems.

First, we need to put the set of numbers in order -->

from: 34, 37, 39, 32, 48, 45, 53, 62, 58, 61, 60, 41 -->

to: 32, 34, 37, 39, 41, 45, 48, 53, 58, 60, 61, 62

maximum = biggest number => thus, 62

median = middle number in a sense => (45+48)/2 => thus, 46.5

upper quartile = median over the median => thus, 60

lower quartile = median under the median => thus, 37

minimum = lowest number => thus, 32

And there we have our 5 answers.

Hope this helps!

PLSSSS!!! (10points)

Answers

Answer:

angle B is 62 Degress angle A is 87 degress D is 87 degress C is 28 degress.

Step-by-step explanation:

I am in geometry btw so i know this stuff and 65 plus 28 is 93 and 180 -93 is 87 so a is 87 and d is 87 too becuase of vertical angles and b is 62 becuase 90 -28 is 62 and c is 28 becuase of vertical angles your wellcome kid good luck!!!!

The mean weight of newborn infants at a community hospital is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. Does the sample data show a significant increase in the average birthrate at a 5% level of significance?
A. Fail to reject the null hypothesis and conclude the mean is 6.6 lb.
B. Reject the null hypothesis and conclude the mean is lower than 6.6 lb.
C. Reject the null hypothesis and conclude the mean is greater than 6.6 lb.
D. Cannot calculate because the population standard deviation is unknown

Answers

Answer:

The correct option is  A

Step-by-step explanation:

From the question we are told that

    The  population is  [tex]\mu = 6.6[/tex]

     The level of significance is [tex]\alpha = 5\% = 0.05[/tex]

      The sample data is  9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds

The Null hypothesis is [tex]H_o : \mu = 6.6[/tex]

 The Alternative hypothesis is  [tex]H_a : \mu > 6.6[/tex]

The critical value of the level of significance obtained from the normal distribution table is

                       [tex]Z_{\alpha } = Z_{0.05 } = 1.645[/tex]

Generally the sample mean is mathematically evaluated as

      [tex]\=x = \frac{\sum x_i }{n}[/tex]

substituting values

      [tex]\=x = \frac{9.0 + 7.3 + 6.0+ 8.8+ 6.8+ 8.4+6.6 }{7}[/tex]

      [tex]\=x = 7.5571[/tex]

The standard deviation is mathematically evaluated as

           [tex]\sigma = \sqrt{\frac{\sum [ x - \= x ]}{n} }[/tex]

substituting values

          [tex]\sigma = \sqrt{\frac{ [ 9.0-7.5571]^2 + [7.3 -7.5571]^2 + [6.0-7.5571]^2 + [8.8- 7.5571]^2 + [6.8- 7.5571]^2 + [8.4 - 7.5571]^2+ [6.6- 7.5571]^2 }{7} }[/tex][tex]\sigma = 1.1774[/tex]

Generally the test statistic is mathematically evaluated as

            [tex]t = \frac{\= x - \mu } { \frac{\sigma }{\sqrt{n} } }[/tex]

substituting values

           [tex]t = \frac{7.5571 - 6.6 } { \frac{1.1774 }{\sqrt{7} } }[/tex]

            [tex]t = 1.4274[/tex]

Looking at the value of  t and  [tex]Z_{\alpha }[/tex]   we see that [tex]t < Z_{\alpha }[/tex] hence we fail to reject the null hypothesis

  What this implies is that there is no sufficient evidence to state that the sample data show as significant increase in the average birth rate

The conclusion is that the mean is  [tex]\mu = 6.6 \ lb[/tex]

Find the minimum sample size n needed to estimate for the given values of​ c, ​, and E. c​, ​, and E Assume that a preliminary sample has at least 30 members.

Answers

Answer:

hello your question is incomplete below is the complete question

Find the minimum sample size n needed to estimate μ For the given values of​ c, σ​, and E. c=0.98​, σ=6.5​, and E=22 Assume that a preliminary sample has at least 30 members.

Answer : 48

Step-by-step explanation:

Given data:

E = 2.2,

std ( σ ) = 6.5

c ( level of confidence ) = 0.98

To find the minimum sample size

we have to first obtain the value of  [tex]Z_{a/2}[/tex]  

note : a can be found using this relation :

( 1 - a ) = 0.98 ----- equation 1

a = 1 - 0.98 = 0.02

hence:  a/2 = 0.01

This means that P( Z ≤ z ) = 0.99  the value of z can be found using the table of standard normal distribution. from the table the value of z = 2.33

P( Z ≤ 2.33 ) = 0.99

To obtain the sample size n

[tex]n = (\frac{std*z}{E} )^{2}[/tex]

n = [tex](\frac{6.5*2.33}{2.2} )^2[/tex] =  (6.88409)^2

Therefore n ≈ 48

Height of a tree increases by 2.5 feet each growing season. Quadratic, linear or exponential?

Answers

Answer:

Linear

Step-by-step explanation:

Given

Height of a tree grows by 2.5 feet

Required

Determine the type of relationship

Take for instance, the height of the tree at year 1 is x

At year 2, it will be x + 2 * 1

At year 3, it will be x + 2 * 2

At year 4, it will be x + 2 * 3

Following same pattern

At year n, it will be x + 2 *(n - 1)

Hence, growth rate = x + 2(n -1)

From the list of given options, the correct answer is Linear because the derived formula above is an example of a linear equation

cSuppose you are standing such that a 45-foot tree is directly between you and the sun. If you are standing 200 feet away from the tree and the tree casts a 225-foot shadow, how tall could you be and still be completely in the shadow of the tree? x 225 ft 200 ft 45 ft Your height is ft (If needed, round to 1 decimal place.)

Answers

Answer:

you could stand at 5.0 ft and still be completely in the shadow of the tree

Step-by-step explanation:

From the diagram attached below;

We consider;

[tex]\overline {BC}[/tex] to be the height of the tree and [tex]\overline {DE}[/tex] to be the height of how tall you could be and still be completely in the shadow of the tree.

∠D = ∠B = 90°

Also;

ΔEAD = ΔBAC   (similar triangles)

Therefore, their sides will also be proportional

i.e

[tex]\dfrac{\overline {DE}}{ \overline {BC}}= \dfrac{\overline{AD}}{ \overline{AC}}[/tex]

[tex]\dfrac{x}{ 45}= \dfrac{225-220}{225}[/tex]

[tex]\dfrac{x}{ 45}= \dfrac{25}{225}[/tex]

By cross multiply

225x = 45 × 25

[tex]x = \dfrac{45 \times 25}{225}[/tex]

[tex]x = \dfrac{1125}{225}[/tex]

x = 5.0 ft

Therefore, you could stand at 5.0 ft and still be completely in the shadow of the tree

distance between 2,-5 and 3,-7

Answers

Answer:

√5

Step-by-step explanation:

[tex](2 ,-5) = (x_1,y_1)\\(3,-7)=(x_2,y_2)\\\\d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ \\d = \sqrt{(3-2)^2 +(-7-(-5))^2}\\ \\d = \sqrt{(1)^2+(-7+5)^2}\\ \\d = \sqrt{(1)^2 + (-2)^2}\\ \\d = \sqrt{1 +4}\\ \\d = \sqrt{5}[/tex]

which statement correctly describes the relation between the variable in the equation C = nd

Answers

Answer:

nd is c

Step-by-step explanation:

Records indicate that x years after 2008, the average property tax on a three bedroom home in a certain community was T(x) =20x^2+40x+600 dollars.

Required:
a. At what rate was the property tax increasing with respect to time in 2008?
b. By how much did the tax change between the years 2008 and 2012?

Answers

Answer:

a) 40 dollars

b) 480 dollars

Step-by-step explanation:

Given the average property tax on a three bedroom home in a certain community modelled by the equation T(x) =20x²+40x+600, the rate at which the property tax is increasing with respect to time in 2008 can be derived by solving for the function T'(x) at x=0

T'(x) = 2(20)x¹ + 40x° + 0

T'(x) = 40x+40

At x = 0,

T'(0) = 40(0)+40

T'(0) = 40

Hence the property tax was increasing at a rate of 40dollars with respect to the initial year (2008).

b) There are 4 years between 2008 and 2012. To know how much that the tax change between the years 2008 and 2012, we will find T(4) - T(0)

Given T(x) =20x²+40x+600

T(4) =20(4)²+40(4)+600

T(4) = 320+160+600

T(4) = 1080 dollars

Also T(0) =20(0)²+40(0)+600

T(0) = 0+0+600

T(0)= 600 dollars

T(4) - T(0) = 1080 - 600

T(4) - T(0) = 480 dollars

Hence, the tax has changed by $480 between 2008 and 2012

Given that
[tex]\sqrt{2p-7}=3[/tex]
and
[tex]7\sqrt{3q-1}=2[/tex]
Evaluate
[tex]p + {q}^{2} [/tex]​

Answers

Answer:

Below

Step-by-step explanation:

The two given expressions are:

● √(2p-7) = 3

● 7√(3q-1) = 2

We are told to evaluate p+q^2

To do that let's find the values of p and q^2

■■■■■■■■■■■■■■■■■■■■■■■■■■

Let's start with p.

● √(2p-7) = 3

Square both sides

● (2p-7) = 3^2

● 2p-7 = 9

Add 7 to both sides

● 2p-7+7 = 9+7

● 2p = 16

Divide both sides by 2

● 2p/2 = 16/2

● p = 8

So the value of p is 8

■■■■■■■■■■■■■■■■■■■■■■■■■■

Let's find the value of q^2

● 7√(3q-1) = 2

Square both sides

● 7^2 × (3q-1) = 2^2

● 49 × (3q-1) = 4

● 49 × 3q - 49 × 1 = 4

● 147q - 49 = 4

Add 49 to both sides

● 147q -49 +49 = 4+49

● 147q = 53

Divide both sides by 147

● 147q/147 = 53/147

● q = 53/ 147

Square both sides

● q^2 = 53^2 / 147^2

● q^2 = 2809/21609

■■■■■■■■■■■■■■■■■■■■■■■■■

● p+q^2 = 8 +(2809/21609)

● p+q^2 = (2809 + 8×21609)/21609

● p+q^2 = 175681 / 21609

● p + q^2 = 8.129

Round it to the nearest unit

● p+ q^2 = 8

solve 27 to the power of (2/3)

Answers

Answer:

9

Step-by-step explanation:

[tex]27^{\frac{2}{3}}\\\mathrm{Factor\:the\:number:\:}\:27=3^3\\=\left(3^3\right)^{\frac{2}{3}}\\\mathrm{Apply\:exponent\:rule}:\\\\\quad \left(a^b\right)^c=a^{bc},\:\quad \:a\ge 0\\\\\left(3^3\right)^{\frac{2}{3}}=3^{3}\times \frac{2}{3}}\\\\3\=times \frac{2}{3}=2\\\\=3^2 \\\\=9[/tex]

[tex]27^{2/3}=(3^3)^{2/3}=3^2=9[/tex]

Julissa gave out an equal number of oranges to each of the 6 apartments on her floor. if she gave each apartment 5 oranges, how many oranges did Julissa give out in all?

Answers

julissa gave equal oranges in 6 apartments

she gave each apartment 5 oranges

so total no. of oranges are = 6×5 = 30

Answer:

D. 30

Step-by-step explanation:

Kenji earned the test scores below in English class.
79, 91, 93, 85, 86, and 88
What are the mean and median of his test scores?

Answers

Answer:

mean=87

median=87

Step-by-step explanation:

mean=sum of test score/number of subject

mean=79+91+93+85+86+88/6

mean=522/6

mean=87

Literal meaning of median is medium.

To find the number which lies in the medium, we must rearrange the number in ascending.

79, 91, 93, 85, 86, 88

79, 85, 86, 88, 91, 93

86+88/2=87

Hope this helps ;) ❤❤❤

Let me know if there is an error in my answer.

According to the Federal Communications Commission, 70% of all U.S. households have vcrs. In a random sample of 15 households, what is the probability that fewer than 13 have vcrs?

Answers

Answer:

The probability  is  [tex]P(x < 13) = 0.8732[/tex]

Step-by-step explanation:

From the question we are told that

    The  probability of success is    p = 0.70

     The  sample size is  [tex]n = 15[/tex]

Generally the distribution of U.S. households have vcrs follow a binomial distribution given that there are only two outcome (household having vcrs or household not having vcrs )

The probability of failure is mathematically evaluated as

       [tex]q = 1- p[/tex]

substituting values

      [tex]q = 1- 0.70[/tex]

      [tex]q = 0.30[/tex]

The probability that fewer than 13 have vcrs is mathematically represented as

          [tex]P(x < 13) = 1- [P(13) + P(14) + P(15)][/tex]

=>     [tex]P(x < 13) = 1-[( \left 15 } \atop {}} \right. C_{13} *p^{13}* q^{15-13})+ (\left 15 } \atop {}} \right. C_{14} *p^{14}* q^{15-14}) +( \left 15 } \atop {}} \right. C_{15} *p^{15}* q^{15-15}) ][/tex]

 Here  [tex]\left 15 } \atop {}} \right. C_{13}[/tex] means  15 combination 13 and the value is  105 (obtained from calculator)

 Here  [tex]\left 15 } \atop {}} \right. C_{14}[/tex] means  15 combination 14 and the value is  15 (obtained from calculator)

 

 Here  [tex]\left 15 } \atop {}} \right. C_{15}[/tex] means  15 combination 15 and the value is  1 (obtained from calculator)

So

 [tex]P(x < 13) = 1-[(105 *p^{13}* q^{2})+ (15 *p^{14}* q^{1}) +(1*p^{15}* q^{0}) ][/tex]

substituting values      

 [tex]P(x < 13) = 1-[(105 *(0.70)^{13}* (0.30)^{2})+ (15 *(0.70)^{14}* (0.30)^{1}) +(1*(0.70)^{15}* (0.30)^{0}) ][/tex]

 [tex]P(x < 13) = 0.8732[/tex]

     

Given the function, Calculate the following values:

Answers

Answer:

[tex]f(-2)=33\\f(-1)=12\\f(0)=1\\f(1)=0\\f(2)=9[/tex]

Step-by-step explanation:

[tex]f(x)=5x^{2} -6x+1\\f(-2)=5(-2)^{2} -6(-2)+1\\f(-2)=5(4)+12+1\\f(-2)=20+13\\f(-2)=33[/tex]

[tex]f(x)=5x^{2}-6x+1\\f(-1)=5(-1)^{2} -6(-1)+1\\f(-1)=5(1)+6+1\\f(-1)=5+7\\f(-1)=12[/tex]

[tex]f(x)=5x^{2}-6x+1\\f(0)=5(0)^{2}-6(0)+1\\f(0)=5(0)-0+1\\f(0)=0+1\\f(0)=1[/tex]

[tex]f(x)=5x^{2}-6x+1\\f(1)=5(1)^{2}-6(1)+1\\f(1)=5(1)-6+1\\f(1)=5-5\\f(1)=0[/tex]

[tex]f(x)=5x^{2}-6x+1\\f(2)=5(2)^{2}-6(2)+1\\f(2)=5(4)-12+1\\f(2)=20-11\\f(2)=9[/tex]

Factor 4(20) + 84. 4(20 + 21) 4(21 + 20) 20(4 + 84) 20(4 + 4)

Answers

Answer:

[tex]\huge\boxed{4 ( 20 + 21)}[/tex]

Step-by-step explanation:

4(20) + 84

Resolve Parenthesis

80 + 84

Taking 4 common as both are the multiples of 4

4 ( 20 + 21)

Find the value of x. A: 15 B: 12 C: 10 D: 8

Answers

Answer:

[tex]\boxed{\sf C. \ 10}[/tex]

Step-by-step explanation:

[tex]\sf The \ intersecting \ chord \ theorem \ states \ that \ the \ products[/tex]

[tex]\sf of \ the \ lengths \ of \ the \ line \ segments \ on \ each \ chord \ are \ equal.[/tex]

[tex]NH \times HT = MH \times HY[/tex]

[tex](x+20) \times 8=12 \times 20[/tex]

[tex]\sf Expand \ brackets \ and \ multiply.[/tex]

[tex]8x+160=240[/tex]

[tex]\sf Subtract \ 160 \ from \ both \ sides.[/tex]

[tex]8x+160-160=240-160[/tex]

[tex]8x=80[/tex]

[tex]\sf Divide \ both \ sides \ by \ 8.[/tex]

[tex]\displaystyle \frac{8x}{8} =\frac{80}{8}[/tex]

[tex]x=10[/tex]

The value of x is 10.

We have a circle and inside it two chords MY and NT intersect at point H.

We have to find the value of x in the figure.

What is intersecting chord theorem?

According to the intersecting chord theorem, when two chords say AB and CD intersect at point O, then

AO x OB = CO x OD

Applying the chord intersecting theorem to the figure in the question, we get -

MH x HY = NH x HT

12 x 20 = (x+20) x 8

240 = 8x + 160

8x = 80

x = 10

Hence the value of x is 10.

To solve more questions on Circles and chords, visit the link below -

https://brainly.com/question/15568573

#SPJ5

You look over the songs in a jukebox and determine that you like of the songs. ​(a) What is the probability that you like the next four songs that are​ played? (Assume a song cannot be​ repeated.) ​(b) What is the probability that you do not like the any of the next four songs that are​ played? (Assume a song cannot be​ repeated.) ​(a) The probability that you like the next four songs that are played is nothing. ​(Round to three decimal places as​ needed.) ​(b) The probability that you do not like any of the next four songs that are played is nothing. ​(Round to three decimal places as​ needed.)

Answers

Complete Question

You look over the songs in a jukebox and determine that you like 18 of 59 songs.

(a) What is the probability that you like the next four songs that are played? (Assume a song cannot be repeated) Round to three decimal places as needed)

(b) What is the probability that you do not like the next four songs that are played? (Assume a song cannot be repeated.) Round to three decimal places as needed

Answer:

a

 [tex]P = 0.0067[/tex]

b

  [tex]Q = 0.222[/tex]

Step-by-step explanation:

From the question we are told that

    The  total number of songs is  [tex]n = 59[/tex]

    The  number of songs you liked is [tex]k = 18[/tex]

The probability that you like the next four songs that are played? (Assume a song cannot be repeated) is mathematically represented as

        [tex]P = \frac{ ^{k} C _4 }{ ^{n} C _4}[/tex]

=>     [tex]P = \frac{ ^{18} C _4 }{ ^{59} C _4}[/tex]

Now using a combination calculator

       [tex]P= \frac{ 3060}{ 455126}[/tex]

       [tex]P = 0.0067[/tex]

The probability that you do not like the next four songs that are played? (Assume a song cannot be repeated.) is mathematically evaluated as

     [tex]Q = \frac{ ^{n- k} C _4 }{ ^{n} C _4}[/tex]

=>  [tex]Q = \frac{ ^{59- 18} C _4 }{ ^{n} C _4}[/tex]

=>  [tex]Q = \frac{ ^{41} C _4 }{ ^{59} C _4}[/tex]

Now using a combination calculator

      [tex]Q = \frac{ 101270}{ 455126}[/tex]

      [tex]Q = 0.222[/tex]

Daniel and Jack together sell 96 tickets to a raffle. Daniel sold 12 more tickets than his friend. How many raffle tickets each friend sell?

Answers

Answer:

Daniel sold 54 and Jack sold 42

Step-by-step explanation:

D = number of tickets that Daniel sold

J = number of tickets that Jack sold

D + J = 96

D = 12+ J

Substitute the second equation into the first equation

12 + J + J = 96

Combine like terms

12 + 2J = 96

Subtract 12 from each side

2J = 84

Divide by 2

J = 42

D = J+12

D = 54

Daniel sold 54 and Jack sold 42

Answer:

Jack sold 42 & Daniel sold 54.

Step-by-step explanation:

96 - 12 = 84

84 / 2 = 42

Jack sold 42.

42 + 12 = 54

Daniel sold 54.

42 + 54 = 96

Hey market sales six cans of food for every seven boxes of food the market sold a total of 26 cans and boxes today how many of each kind did the market sale

Answers

Answer:

It sold 14 cans boxes of food and 12 cans of food.

Step-by-step explanation:

The factor for the food cans depend upon every seven food boxes .So, the same no. of sets of food cans will be sold.

Let the no. of sets of food boxes be x.

According to the question,

6x+7x=26

13x=26

x=26/13

x=2

No. of food cans =6x=6×2=12 cans

No. of food boxes=7x=7×2=14 boxes

Please mark brainliest ,if it is truly the best ! Thank you!

It has been reported that 20.4% of incoming freshmen indicate that they will major in business or a related field. A random sample of 400 incoming college freshmen was asked their preference, and 95 replied that they were considering business as a major. Estimate the true proportion of freshman business majors with 98% confidence. Does your interval contain 20.4%?

Answers

Answer:

The  98% confidence interval

                         [tex]0.1884 < p < 0.2876[/tex]

The confidence interval contains  20.4%

Step-by-step explanation:

From the question we are told that

    The sample size is  n =  400

The number that replied that they were considering business as a major [tex]x = 95[/tex]

  The  sample proportion is mathematically  evaluated as

          [tex]\r p = \frac{95}{400}[/tex]

         [tex]\r p = 0.238[/tex]

Given that the confidence level 98% then the level of significance is evaluated as

      [tex]\alpha = 100 - 98[/tex]

     [tex]\alpha = 2 \%[/tex]

     [tex]\alpha = 0.02[/tex]

Next we obtain the critical value of [tex]\frac{ \alpha }{2}[/tex]  from the normal distribution table is  

       [tex]Z_{\frac{ \alpha }{2} } = 2.33[/tex]

  Generally the margin of error is mathematically represented as

       [tex]E = Z_{\frac{ \alpha }{2} } * \sqrt{ \frac{ p (1 - p )}{n} }[/tex]  

       [tex]E = 2.33 * \sqrt{ \frac{ 0.238 (1 - 0.238 )}{400} }[/tex]

        [tex]E = 0.0496[/tex]

The  98%  confidence interval is mathematically represented

   [tex]\r p - E < p < \r p + E[/tex]

 =>    [tex]0.238 - 0.0496 < p <0.238 + 0.0496[/tex]

=>      [tex]0.1884 < p < 0.2876[/tex]

Given the number of trials and the probability of success, determine the probability indicated: a. n = 15, p = 0.4, find P(4 successes) b. n = 12, p = 0.2, find P(2 failures) c. n = 20, p = 0.05, find P(at least 3 successes)

Answers

Answer:

A)0.126775 B)0.000004325376 C) 0.07548

Step-by-step explanation:

Given the following :

A.) a. n = 15, p = 0.4, find P(4 successes)

a = number of trials p=probability of success

P(4 successes) = P(x = 4)

USING:

nCx * p^x * (1-p)^(n-x)

15C4 * 0.4^4 * (1-0.4)^(15-4)

1365 * 0.0256 * 0.00362797056

= 0.126775

B)

b. n = 12, p = 0.2, find P(2 failures),

P(2 failures) = P(12 - 2) = p(10 success)

USING:

nCx * p^x * (1-p)^(n-x)

12C10 * 0.2^10 * (1-0.2)^(12-10)

66 * 0.0000001024 * 0.64

= 0.000004325376

C) n = 20, p = 0.05, find P(at least 3 successes)

P(X≥ 3) = p(3) + p(4) + p(5) +.... p(20)

To avoid complicated calculations, we can use the online binomial probability distribution calculator :

P(X≥ 3) = 0.07548

Which of the following is equal to the rational expression below when x=-1
or -8?
11(x+8)
/(x + 1)(x+8)​

Answers

Answer:

11/(x + 1) thus d: is the answer

Step-by-step explanation:

Simplify the following:

(11 (x + 8))/((x + 1) (x + 8))

(11 (x + 8))/((x + 1) (x + 8)) = (x + 8)/(x + 8)×11/(x + 1) = 11/(x + 1):

Answer: 11/(x + 1)


Help please!!! Thank you

Answers

Answer:

5/7

Step-by-step explanation:

There are a couple ways to solve this.  One would be by finding the least common denominator for each one with 2/3, subtracting, and seeing what is left over.  Another way is converting to decimals.

2/3=0.666666

————————-

7/8=0.875

8/9=0.88888

4/5=0.8

5/7=0.7143

They are all greater than 2/3 (0.6666666), but 5/7 is the closest, so would have the least waste.

Evaluate 2/3 + 1/3 + 1/6 + …

Answers

Answer:

7/6

Step-by-step explanation:

The LCD of these three fractions is 6; the denominators 3, 3 and 6 divide evenly into 6.

Therefore we have:

4/6 + 2/6 + 1/6 = 7/6

Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 35.0 hours and a standard deviation of 5.5 hours. As a part of its quality assurance program, Power +, Inc. tests samples of 25 batteries.
A) What can you say about the shape of the distribution of the sample mean?
B) What is the standard error of the distribution of the sample mean?
C) What proportion of the samples will have a mean useful life of more than 36 hours?
D) What proportion of the sample will have a mean useful life greater than 34.5 hours?
E) What proportion of the sample will have a mean useful life between 34.5 and 36.0 hours?

Answers

Answer:

(A) The shape of the distribution of the sample mean is bell-shaped.

(B) The standard error of the distribution of the sample mean is 1.1.

(C) The proportion of the samples that have a mean useful life of more than 36 hours is 0.1814.

(D) The proportion of the sample that has a mean useful life greater than 34.5 hours is 0.6736.

(E) The proportion of the sample that has a mean useful life between 34.5 and 36.0 hours is 0.4922.

Step-by-step explanation:

We are given that Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 35.0 hours and a standard deviation of 5.5 hours.

As a part of its quality assurance program, Power +, Inc. tests samples of 25 batteries.

Let [tex]\bar X[/tex] = sample mean life of these batteries

(A) The shape of the distribution of the sample mean will be bell-shaped because the sample mean also follows the normal distribution as it is taken from the population data only.

(B) The standard error of the distribution of the sample mean is given by;

            Standard error =  [tex]\frac{\sigma}{\sqrt{n} }[/tex]

Here, [tex]\sigma[/tex] = standard deviation = 5.5 hours

         n = sample of batteries = 25

So, the standard error =  [tex]\frac{5.5}{\sqrt{25} }[/tex]  = 1.1.

(C) The z-score probability distribution for the sample mean is given by;

                               Z  =  [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = population mean life of battery = 35.0 hours

            [tex]\sigma[/tex] = standard deviation = 5.5 hours

            n = sample of batteries = 25

Now, the proportion of the samples that will have a mean useful life of more than 36 hours is given by = P([tex]\bar X[/tex] > 36 hours)

     

       P([tex]\bar X[/tex] > 36 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{36-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z > 0.91) = 1 - P(Z [tex]\leq[/tex] 0.91)

                                                               = 1 - 0.8186 = 0.1814

(D) The proportion of the samples that will have a mean useful life of more than 34.5 hours is given by = P([tex]\bar X[/tex] > 34.5 hours)

     

       P([tex]\bar X[/tex] > 34.5 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{34.5-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z > -0.45) = P(Z [tex]\leq[/tex] 0.45)

                                                                    = 0.6736

(E) The proportion of the samples that will have a mean useful life between 34.5 and 36.0 hours is given by = P(34.5 hrs < [tex]\bar X[/tex] > 36 hrs)

     P(34.5 hrs < [tex]\bar X[/tex] < 36 hrs) = P([tex]\bar X[/tex] < 36 hrs) - P([tex]\bar X[/tex] [tex]\leq[/tex] 34.5 hrs)

     P([tex]\bar X[/tex] < 36 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{36-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z < 0.91) = 0.8186

     P([tex]\bar X[/tex] [tex]\leq[/tex] 34.5 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{34.5-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z [tex]\leq[/tex] -0.45) = 1 - P(Z [tex]\leq[/tex] 0.45)

                                                                    = 1 - 0.6736 = 0.3264                              

Therefore, P(34.5 hrs < [tex]\bar X[/tex] < 36 hrs) = 0.8186 - 0.3264 = 0.4922.

When is it easier to use the addition method rather than the substitution method to solve a system of equations?

Answers

Answer: When the addition of two or more equations leads to the elimination of one of the variables.

Step-by-step explanation:

When we have a system of equations, the addition method seems to be useful only when adding the equations will lead to the elimination of one of the variables:

An example of this can be, for the variables x and y:

3*x + x*y - 2*y = 3

x^2 + x*y - 2y = 42

now we can "add" (actually subtract) the equations and get (eq2 minus eq1)

(x^2 + x*y - 2y) - (3*x + x*y - 2*y ) = 42 - 3

x^2 - 3*x = 39

x^2 - 3*x - 39 = 0

And now we can solve it for x, and then find the value of y.

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