Josh sees a pair of trainers with a tags saying 75% off the recommended price of £80. Josh decided to buy the trainers. How many pounds will it cost him?

Answers

Answer 1
To find this use this formula: (percent off) / 100 x (cost)

Related Questions

Add :-

a+2b-3c, -3a+b+2cand 2a -3b+c​

Answers

Answer:

[tex]a + 2b -3 c + - 3a + b + 2c + 2a - 3b + c \\ = a - 3a + 2a + 2b + b - 3b - 3c + 2c + c \\ 0a + 0b + 0c \\ thank \: you[/tex]

a+2b-3c+(-3a+b+2c) +(2a-3b+c)

=a+2b-3c-3a+b+2c+2a-3b+c

=a-3a+2b+2b+b-3b-3c++2b+c

=0a+0b+0c

=0

Therefore, the addition of the expressions, a+2b-3c+(-3a+b+2c) +(2a-3b+c) is zero or 0.

To know more about algebraic expressions

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Use the tables to answer the question.
How can you determine which store offers a better price on guitar lessons?
A. You cannot determine this from the information given.
B. Compare the most expensive price from each store’s table.
C. Compare the least expensive price from each store’s table.
D. Find the cost of 1 week of guitar lessons for each store and compare.

Answers

Answer:

D. Find the cost of 1 week of guitar lessons for each store and compare.

Step-by-step explanation:

Find the cost for 1 week of guitar lessons by finding the slope of the data.

[tex]\frac{y_2-y_1}{x_2-x_1}\\\\ A.)\frac{132-66}{6-3}=\frac{66}{3}=22\\\\ B.)\frac{168-84}{8-4}=\frac{84}{4}=21[/tex]

Therefore, Store B would be the best choice.

4. Manuel swims at a speed of 1 yard per second. How many feet per minute does he swim?​

Answers

Answer:

180 Feet Per Minute

Step-by-step explanation:

Please mark me as brainliest and rate 5 stars!

180 feet per minute.

Evaluate:
[tex]{ \int \limits^\pi_{ \frac{1}{4}\pi}{ {e {}^{2 \sigma} (\sqrt{1 - { \sigma}^{2} } ) d \sigma}}}[/tex]

Answers

Answer:

hope this answer helps.

QUESTION 3.1 POINT
An investment pays 25% interest compounded monthly. What percent, as a decimal, is the effective annual yied? Enter your
answer as a decimal rounded to four decimal places.

Answers

9514 1404 393

Answer:

  0.2807

Step-by-step explanation:

The relationship between the effective annual yield (e) and the nominal annual interest rate (r) compounded n times per year is ...

  e = (1 +r/n)^n -1

  e = (1 +0.25/12)^12 -1 = 0.2807 . . . . . . about 28.07%

The ratio of frogs to toads was 3 to 7. If there were 1280 frogs and toads in all, how many were frogs?

Answers

Answer:

348 frogs

Step-by-step explanation:

ratio = 3:7

total of ratio = 10

frogs = 3/10 × 1280 = 348 frogs

Answer:

let the ratio be 3x and 7x.

3x+7x=1280

10x=1280

x=128

Now

frogs =3x=3*128=384

toads =7x=7*128=896

If lan does a job in 132 hours and with the help of Danielle they can do it together in 44 hours, how long
would it take Danielle to do it alone

Answers

Answer:

88 hours

Step-by-step explanation:

Ian = 132

Daniel + Ian = 44

Daniel = Ian - 44

= 132-44 = 88 hours

The resistors produced by a manufacturer are required to have an average resistance of 0.150 ohms. Statistical analysis of the output suggests that the resistances can be approximated by a normal distribution with known standard deviation of 0.005 ohms. We are interested in testing the hypothesis that the resistors conform to the specifications.

Requied:
a. Determine whether a random sample of 10 resistors yielding a sample mean of 0.152 ohms indicates that the resistors are conforming. Use alpha = 0.05.
b. Calculate a 95% confidence interval for the average resistance. How does this interval relate to your solution of part (a)?

Answers

Answer:

a) The p-value of the test is 0.2076 > 0.05, which means that the sample indicates that the resistors are conforming.

b) The 95% confidence interval for the average resistance is (0.147, 0.153). 0.152 is part of the confidence interval, which means that as the test statistic in item a), it indicates that the resistors are conforming.

Step-by-step explanation:

Question a:

The resistors produced by a manufacturer are required to have an average resistance of 0.150 ohms.

At the null hypothesis, we test if this is the average resistance, that is:

[tex]H_0: \mu = 0.15[/tex]

We are interested in testing the hypothesis that the resistors conform to the specifications.

At the alternative hypothesis, we test if it is not conforming, that is, the mean is different of 0.15, so:

[tex]H_1: \mu \neq 0.15[/tex]

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.

0.15 is tested at the null hypothesis:

This means that [tex]\mu = 0.15[/tex]

Sample mean of 0.152, sample of 10, population standard deviation of 0.005.

This means that [tex]X = 0.152, n = 10, \sigma = 0.005[/tex]

Value of the test statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{0.152 - 0.15}{\frac{0.005}{\sqrt{10}}}[/tex]

[tex]z = 1.26[/tex]

P-value of the test and decision:

The p-value of the test is the probability of the sample mean differing from 0.15 by at least 0.152 - 0.15 = 0.002, which is P(|z| > 1.26), given by two multiplied by the p-value of z = -1.26.

Looking at the z-table, z = -1.26 has a p-value of 0.1038.

2*0.1038 = 0.2076

The p-value of the test is 0.2076 > 0.05, which means that the sample indicates that the resistors are conforming.

Question b:

We have to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a p-value of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 1.96\frac{0.005}{\sqrt{10}} = 0.003[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is 0.15 - 0.003 = 0.147.

The upper end of the interval is the sample mean added to M. So it is 0.15 + 0.003 = 0.153.

The 95% confidence interval for the average resistance is (0.147, 0.153). 0.152 is part of the confidence interval, which means that as the test statistic in item a), it indicates that the resistors are conforming.

There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean 3 cm and standard deviation 0.10 cm. The second machine produces corks with diameters that have a normal distribution with mean 3.04 cm and standard deviation 0.04 cm. Acceptable corks have diameters between 2.9 cm and 3.1 cm. What is the probability that the first machine produces an acceptable cork

Answers

Answer:

0.6827

Step-by-step explanation:

Given that :

Mean, μ = 3

Standard deviation, σ = 0.1

To produce an acceptable cork. :

P(2.9 < X < 3.1)

Recall :

Z = (x - μ) / σ

P(2.9 < X < 3.1) = P[((2.9 - 3) / 0.1) < Z < ((3.1 - 3) / 0.1)]

P(2.9 < X < 3.1) = P(-1 < Z < 1)

Using a normal distribution calculator, we obtain the probability to the right of the distribution :

P(2.9 < X < 3.1) = P(1 < Z < - 1) = 0.8413 - 0.1587 = 0.6827

Hence, the probability that the first machine produces an acceptable cork is 0.6827

-5 + 3 and also what is 1/4 of 24
What is the answer i am struggling

Answers

Answer:

-5+3=-2

1/4 of 24 = 6

Step-by-step explanation:

PLEASEE HELP ME ASAPPP (geometry)

Answers

Answer:AE=EC và BF=FC => EF là đường trung bình của tam giác ABC

=> EF // và bằng 1/2 AB

=> AB = 16

Step-by-step explanation:

Answer:

AB=16

Step-by-step explanation:

Midsegment Theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long.

The mid-segment of a triangle, which joins the midpoints of two sides of a triangle, is parallel to the third side of the triangle and half the length of that third side of the triangle.

AD=DB

AD+DB=AB=2EF

AB=2×8=16

3 write the factor of the following (1) 48 (2) 36 (3) 28 (4) 100 (5) 125 ​

Answers

Answer:

FACTORING THE NUMBERS :-

well u didnt say to prime factors so i am writing all factors

1) 48 => 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48    ( all are plus and minus )

2) 36 => 1, 2, 3, 4, 6, 9, 12, 18 and 36    ( all are plus and minus )

3) 28 => 1, 2, 4, 7, 14 and 28    ( all are plus and minus )

4) 100 => 1, 2, 4, 5, 10, 20, 25, 50, and 100  ( all are plus and minus )

5) 125 => 1, 5, 25, 125   ( all are plus and minus )

is it worth brainliest...

yes ofc

Use the commutative law of multiplication to rewrite 67 x 13.

A. 3 X 671
B. 13 x 67
C.6 X 7 X1 X3
D.80

Answers

Answer:

A. 671*3

B. 67*13

C. 3*1*7*6

D. 1*80

The PTA sells 100 tickets for a raffle and puts them in a bowl. They will randomly pull out a ticket for the first prize and then another ticket for the second prize. You have 10 tickets and your friend has 10 tickets. What is the probability that your friend wins the first prize and you win the second prize?

Answers

Jÿïôò śfrtÿ hjkÿï èrï

[tex]\sqrt{25}[/tex]

Answers

Answer:

5

Step-by-step explanation:

Calculate the square root of 25 and get 5.

5 I just thought of what # gets me to 25

Police sometimes measure shoe prints at crime scenes so that they can learn something about criminals. Listed below are shoe print​ lengths, foot​ lengths, and heights of males. Construct a​ scatterplot, find the value of the linear correlation coefficient​ r, and find the​ P-value of r. Determine whether there is sufficient evidence to support a claim of linear correlation between the two variables. Based on these​ results, does it appear that police can use a shoe print length to estimate the height of a​ male? Use a significance level of α=0.01

Answers

It does not appear that police can use a shoe print length to estimate the height of a​ male.

The given parameters are:

[tex]\begin{array}{cccccc}{Shoe\ Print} & {28.6} & {29.4} & {32.2} & {32.4} & {27.3} \ \\ Height (cm) & {172.5} & {176.7} & {188.4} & {170.1} & {179.2} \ \end{array}[/tex]

Rewrite as:

[tex]\begin{array}{cccccc}{x} & {28.6} & {29.4} & {32.2} & {32.4} & {27.3} \ \\ y & {172.5} & {176.7} & {188.4} & {170.1} & {179.2} \ \end{array}[/tex]

See attachment for scatter plot

To determine the correlation coefficient, we extend the table as follows:

[tex]\begin{array}{cccccc}{x} & {28.6} & {29.4} & {32.2} & {32.4} & {27.3} & y & {172.5} & {176.7} & {188.4} & {170.1} & {179.2} & x^2 & {817.96} & {864.36} & {1036.84} & {1049.76} & {745.29} & y^2 & {29756.25} & {31222.89} & {35494.56} & {28934.01} & {32112.64} & x \times y & {4933.5} & {5194.98} & {6066.48} & {5511.24} & {4892.16} \ \end{array}[/tex]

The correlation coefficient (r) is:

[tex]r = \frac{\sum(x - \bar x)(y - \bar y)}{\sqrt{SS_x * SS_y}}[/tex]

We have:

[tex]n =5[/tex]

[tex]\sum xy =4933.5+5194.98+6066.48+5511.24+4892.16 =26598.36[/tex]

[tex]\sum x =28.6+29.4+32.2+32.4+27.3=149.9[/tex]

[tex]\sum y =172.5+176.7+188.4+170.1+179.2=886.9[/tex]

[tex]\sum x^2 =817.96+864.36+1036.84+1049.76+745.29=4514.21[/tex]

[tex]\sum y^2 =29756.25+31222.89+35494.56+28934.01+32112.64=157520.35[/tex]

Calculate mean of x and y

[tex]\bar x = \frac{\sum x}{n} = \frac{149.9}{5} = 29.98[/tex]

[tex]\bar y = \frac{\sum y}{n} = \frac{886.9}{5} = 177.38[/tex]

Calculate SSx and SSy

[tex]SS_x = \sum (x - \bar x)^2 =(28.6-29.98)^2 + (29.4-29.98)^2 + (32.2-29.98)^2 + (32.4-29.98)^2 + (27.3-29.98)^2 =20.208[/tex]

[tex]SS_y = \sum (y - \bar x)^2 =(172.5-177.38)^2 + (176.7-177.38)^2 + (188.4-177.38)^2 + (170.1-177.38)^2 + (179.2-177.38)^2 =202.028[/tex]

Calculate [tex]\sum(x - \bar x)(y - \bar y)[/tex]

[tex]\sum(x - \bar x)(y - \bar y) = (28.6-29.98)*(172.5-177.38) + (29.4-29.98)*(176.7-177.38) + (32.2-29.98)*(188.4-177.38) + (32.4-29.98)*(170.1-177.38) + (27.3-29.98) *(179.2-177.38) =9.098[/tex]

So:

[tex]r = \frac{\sum(x - \bar x)(y - \bar y)}{\sqrt{SS_x * SS_y}}[/tex]

[tex]r = \frac{9.098}{\sqrt{20.208 * 202.028}}[/tex]

[tex]r = \frac{9.098}{\sqrt{4082.581824}}[/tex]

[tex]r = \frac{9.098}{63.90}[/tex]

[tex]r = 0.142[/tex]

Calculate test statistic:

[tex]t = \frac{r}{\sqrt{\frac{1 - r^2}{n-2}}}[/tex]

[tex]t = \frac{0.142}{\sqrt{\frac{1 - 0.142^2}{5-2}}}[/tex]

[tex]t = \frac{0.142}{\sqrt{\frac{0.979836}{3}}}[/tex]

[tex]t = \frac{0.142}{\sqrt{0.326612}}[/tex]

[tex]t = \frac{0.142}{0.5715}[/tex]

[tex]t = 0.248[/tex]

Calculate the degrees of freedom

[tex]df = n - 2 = 5 - 2 = 3[/tex]

The [tex]t_{\alpha/2}[/tex] value at:

[tex]df =3[/tex]

[tex]t = 0.248[/tex]

[tex]\alpha = 0.01[/tex]

The value is:

[tex]t_{0.01/2} = \±5.841[/tex]

This means that we reject the null hypothesis if the t value is not between -5.841 and 5.841

We calculate the t value as:

[tex]t = 0.248[/tex]

[tex]-5.841 < 0.248 < 5.841[/tex]

Hence, we do not reject the null hypothesis because they do not appear to have any correlation.

Read more about regression at:

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a farmer needs 5 men to clear his farm in 10 days. How many men will he need if he must finish clearing the farm in two days if they work at the same rate?

Answers

Answer:

25 workers

Step-by-step explanation:

If you like my answer than please mark me brainliest thanks

,

pls help me asap !!!

Answers

Answer:

11

Step-by-step explanation:

Hopefully you can see that this is an isosceles triangle and remembering the inequality theorem of a triangle (4,4,11 triangle cannot exist).  Iso triangle has two side the same length - as well as two angles the same.

Solve x∕3 < 5 Question 5 options: A) x ≥ 15 B) x > 15 C) x < 15 D) x ≤ 15

Answers

Answer:

C

Step-by-step explanation:

Given

[tex]\frac{x}{3}[/tex] < 5 ( multiply both sides by 3 to clear the fraction )

x < 15 → C

Doyle Company issued $500,000 of 10-year, 7 percent bonds on January 1, 2018. The bonds were issued at face value. Interest is payable in cash on December 31 of each year. Doyle immediately invested the proceeds from the bond issue in land. The land was leased for an annual $125,000 of cash revenue, which was collected on December 31 of each year, beginning December 31, 2018

Answers

Answer:

f

Step-by-step explanation:

Please help!!

Find BD​

Answers

Answer:  [tex]8\sqrt{2}[/tex]

==========================================================

Work Shown:

Focus entirely on triangle ABD (or on triangle BCD; both are identical)

The two legs of this triangle are AB = 8 and AD = 8. The hypotenuse is unknown, so we'll say BD = x.

Apply the pythagorean theorem.

[tex]a^2 + b^2 = c^2\\\\c = \sqrt{a^2 + b^2}\\\\x = \sqrt{8^2 + 8^2}\\\\x = \sqrt{2*8^2}\\\\x = \sqrt{8^2*2}\\\\x = \sqrt{8^2}*\sqrt{2}\\\\x = 8\sqrt{2}\\\\[/tex]

So that's why the diagonal BD is exactly [tex]8\sqrt{2}\\\\[/tex] units long

Side note: [tex]8\sqrt{2} \approx 11.3137[/tex]

The distance between Ali's house and 1 point
college is exactly 135 miles. If she
drove 2/3 of the distance in 135
minutes. What was her average speed
in miles per hour?

Answers

First we have to figure out how long it would take for the full voyage and that would be 135 + (135 x 1/3) and the answer to that would be 135 + 45 = 180 and that means that 180 is the total minutes it would take to travel the whole trip, now we have to calculate average speed which would be 135(distance)/180(time) which would end up being 135 miles/ 3 hours, now we divide the entire equation by 3 which would be 45/1
CONCLUSION ——————————
Ali would be driving 45 miles per hour

Ali's average speed was 40 miles per hour.

What is an average speed?

The total distance traveled is to be divided by the total time consumed brings us the average speed.

How to calculate the average speed of Ali?

The total distance between the college from Ali's house is 135 miles.

She drove 2/3rd of the total distance in 135 minutes.

She drove =135*2/3miles

=90miles.

Ali can drive 90miles in 135 mins.

Therefore, her average speed is: 90*60/135 miles per hour.

=40 miles per hour.

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190 of 7
6 7 8 9 10
-3
4
5
6
The slope of the line shown in the graph is
and the intercept of the line is

Answers

Answer:slope 2/3

Y-int 6

Step-by-step explanation:

A rectangle has a length of 7 in. and a width of 2 in. if the rectangle is enlarged using a scale factor of 1.5, what will be the perimeter of the new rectangle

Answers

Answer:

27 inch

Step-by-step explanation:

Current perimeter=18

New perimeter=18*1.5=27 in

the number of multiples of a given number is infinite ( )​

Answers

Answer:

make an 8 horizontal

oooookkkk

Answer:

TRUE

The number of multiples of a given number is finite is a false statement. The number of multiples of a given number is infinite.

Examples:

Multiples of 2 = 2,4,6,8,10,…..

Multiples of 3 = 3,6,9,12,15,18,…

Multiples of 4 = 4, 8, 12, 16, 120, 24….

∴ The number of multiples of a given number is infinite .

Answer From Gauth Math

5x-22 3x +105 x minus 22 3 X + 10 ​

Answers

-291x+10

:)))))) Have fun

A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with the mean 3, find the probability that the shop sells. . (a) At least 3 in a week. (b) At most 7 in a week. (c) More than 20 in a month (4 weeks).

Answers

Answer:

a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.

b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.

c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.

Step-by-step explanation:

For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}[/tex]

In which

x is the number of successes

e = 2.71828 is the Euler number

[tex]\lambda[/tex] is the mean in the given interval.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The Poisson distribution can be approximated to the normal with [tex]\mu = \lambda, \sigma = \sqrt{\lambda}[/tex], if [tex]\lambda>10[/tex].

Poisson variable with the mean 3

This means that [tex]\lambda= 3[/tex].

(a) At least 3 in a week.

This is [tex]P(X \geq 3)[/tex]. So

[tex]P(X \geq 3) = 1 - P(X < 3)[/tex]

In which:

[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

Then

[tex]P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498[/tex]

[tex]P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494[/tex]

[tex]P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240[/tex]

So

[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232[/tex]

[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - 0.4232 = 0.5768[/tex]

0.5768 = 57.68% probability that the shop sells at least 3 in a week.

(b) At most 7 in a week.

This is:

[tex]P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)[/tex]

In which

[tex]P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498[/tex]

[tex]P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494[/tex]

[tex]P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240[/tex]

[tex]P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240[/tex]

[tex]P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680[/tex]

[tex]P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008[/tex]

[tex]P(X = 6) = \frac{e^{-3}*3^{6}}{(6)!} = 0.0504[/tex]

[tex]P(X = 7) = \frac{e^{-3}*3^{7}}{(7)!} = 0.0216[/tex]

Then

[tex]P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 + 0.0216 = 0.988[/tex]

0.988 = 98.8% probability that the shop sells at most 7 in a week.

(c) More than 20 in a month (4 weeks).

4 weeks, so:

[tex]\mu = \lambda = 4(3) = 12[/tex]

[tex]\sigma = \sqrt{\lambda} = \sqrt{12}[/tex]

The probability, using continuity correction, is P(X > 20 + 0.5) = P(X > 20.5), which is 1 subtracted by the p-value of Z when X = 20.5.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{20 - 12}{\sqrt{12}}[/tex]

[tex]Z = 2.31[/tex]

[tex]Z = 2.31[/tex] has a p-value of 0.9896.

1 - 0.9896 = 0.0104

0.0104 = 1.04% probability that the shop sells more than 20 in a month.

The probability of the selling the video recorders for considered cases are:

P(At least 3 in a week) = 0.5768 approximately.P(At most 7 in a week) = 0.9881 approximately.P( more than 20 in a month) = 0.0839 approximately.

What are some of the properties of Poisson distribution?

Let X ~ Pois(λ)

Then we have:

E(X) = λ = Var(X)

Since standard deviation is square root (positive) of variance,

Thus,

Standard deviation of X = [tex]\sqrt{\lambda}[/tex]

Its probability function is given by

f(k; λ) = Pr(X = k) = [tex]\dfrac{\lambda^{k}e^{-\lambda}}{k!}[/tex]

For this case, let we have:

X = the number of weekly demand of video recorder for the considered shop.

Then, by the given data, we have:

X ~ Pois(λ=3)


Evaluating each event's probability:

Case 1: At least 3 in a week.

[tex]P(X > 3) = 1- P(X \leq 2) = \sum_{i=0}^{2}P(X=i) = \sum_{i=0}^{2} \dfrac{3^ie^{-3}}{i!}\\\\P(X > 3) = 1 - e^{-3} \times \left( 1 + 3 + 9/2\right) \approx 1 - 0.4232 = 0.5768[/tex]

Case 2: At most 7 in a week.

[tex]P(X \leq 7) = \sum_{i=0}^{7}P(X=i) = \sum_{i=0}^{7} \dfrac{3^ie^{-3}}{i!}\\\\P(X \leq 7) = e^{-3} \times \left( 1 + 3 + 9/2 + 27/6 + 81/24 + 243/120 + 729/720 + 2187/5040\right)\\\\P(X \leq 7) \approx 0.9881[/tex]

Case 3: More than 20 in a month(4 weeks)

That means more than 5 in a week on average.

[tex]P(X > 5) = 1- P(X \leq 5) =\sum_{i=0}^{5}P(X=i) = \sum_{i=0}^{5} \dfrac{3^ie^{-3}}{i!}\\\\P(X > 5) = 1- e^{-3}( 1 + 3 + 9/2 + 27/6 + 81/24 + 243/120)\\\\P(X > 5) \approx 1 - 0.9161 \\ P(X > 5) \approx 0.0839[/tex]


Thus, the probability of the selling the video recorders for considered cases are:

P(At least 3 in a week) = 0.5768 approximately.P(At most 7 in a week) = 0.9881 approximately.P( more than 20 in a month) = 0.0839 approximately.

Learn more about poisson distribution here:

https://brainly.com/question/7879375

Solve. x+y+z=6 3x−2y+2z=2−2x−y+3z=−4

Answers

Answer:

-4?

hope dis helps ^-^

If a normally distributed population has a mean (mu) that equals 100 with a standard deviation (sigma) of 18, what will be the computed z-score with a sample mean (x-bar) of 106 from a sample size of 9?

Answers

Answer:

Z = 1

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean (mu) that equals 100 with a standard deviation (sigma) of 18

[tex]\mu = 100, \sigma = 18[/tex]

Sample of 9:

This means that [tex]n = 9, s = \frac{18}{\sqrt{9}} = 6[/tex]

What will be the computed z-score with a sample mean (x-bar) of 106?

This is Z when X = 106. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{106 - 100}{6}[/tex]

[tex]Z = 1[/tex]

So Z = 1 is the answer.

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Answers

Step-by-step explanation:

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Answer:

Oop

Step-by-step explanation:

I’m bored

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