Answer:
2ml
Step-by-step explanation:
50mg of some potent agent has to be given every 12 hours.
there is a solution that has a concentration of that agent of 125mg/5ml
we need to administer some part of this solution, which we cannot (or should not) change in its structure.
that means the ratio of agent to overall solution stays the same, no matter how much of the solution we administer.
all we need to do is to transit the ratio of 125/5 to represent 50/x (maintaining the said ratio).
in other words, we need to find how many ml we need to administer, so that 50mg of the agent enter the body.
so,
125/5 = 50/x
125x/5 = 50
25x = 50
x = 50/25 = 2
2ml of the solution needs to be administered every 12 hours.
a. Consider the situation where you have three game chips, each labeled with one of the the numbers 3, 5, and 10 in a hat a. If you draw out 2 chips without replacement between each chip draw, list the entire sample space of po ssible results that can occur in the draw Use the three events are defined as follows, to answer parts b through n below:
Event A: the sum of the 2 drawn numbers is even.
Event B: the sum of the 2 drawn numbers is odd.
Event C: the sum of the 2 drawn numbers is a prime number
Now, using your answer to part a find the following probability values
b. P (A)=
c. P (B)=
d. P (C)=
e. P (A and C)-=
f. P(A or B)=
g. P (B andC)=
h. P(A or C)- =
i. P (C given B)=
j. P(C given A)=
k. P (not B)=
l. P (not C)=
Are events A and B mutually exclusive?Why or why not?
Are events B and C mutually exclusive? Why or why not?
Answer:
a) {3,5}{3,10}{5,10}
b) [tex]P(A)=\frac{1}{3}[/tex]
c) [tex]P(B)=\frac{2}{3}[/tex]
d) [tex]P(C)=\frac{1}{3}[/tex]
e) [tex]P(A and C)=0[/tex]
f) [tex]P(A or B)=1[/tex]
g) [tex]P(B and C)=\frac{1}{3}[/tex]
h) [tex]P(A or C)=\frac{2}{3}[/tex]
i) [tex]P(C given B)=\frac{1}{2}[/tex]
j) [tex]P(C given A)=0[/tex]
k) [tex]P(not B)=\frac{1}{3}[/tex]
l) [tex]P(not C)=\frac{2}{3}[/tex]
Yes, events A and B are mutually exclusive. Because the results can either be even or odd, not both. No, events B and C are not mutually exclusive because the result can be both, odd and prime.
Step-by-step explanation:
a)
In order to solve part a of the problem, we need to find the possible outcomes, in this case, the possible outcomes are:
{3,5}{3,10} and {5,10}
We could think of the oppsite order, for example {5,3}{10,3}{10,5} but these are basically the same as the previous outcomes, so we will just take three outcomes in our sample space. We can think of it as drawing the two chips at the same time.
b)
Now the probability of the sum of the chips to be even. There is only one outcome where the sum of the chips is even, {3,5} since 3+5=8 the other outcomes will give us an odd number, so:
[tex]P=\frac{#desired}{#possible}[/tex]
[tex]P(A)=\frac{1}{3}[/tex]
c) For the probability of the sum of the chips to be odd, there are two outcomes where the sum of the chips is odd, {3,10} since 3+10=13 and {5,10} since 5+10=15 the other outcomes will give us an even number, so:
[tex]P(B)=\frac{2}{3}[/tex]
d) The probability of the sum of the chips is prime. There is only one outcome where the sum of the chips is prime, {3,10} since 3+10=13 the other outcomes will give us non prime results, so:
[tex]P(C)=\frac{1}{3}[/tex]
e) The probability of the sum of the chips to be even and prime. There are no results where we can get an even and prime number, since the only even and prime number there is is number 2 and no outcome will give us that number, so:
P(A and C)=0
f) The probability of the sum of the chips is even or odd. We can either get even or odd results, so no matter what outcome we get, we will get an odd or even result so:
[tex]P(A or B)=1[/tex]
g) The probability of the sum of the chips is odd and prime. There is only one outcome where the sum of the chips is odd and prime, {3,10} since 3+10=13 the other outcomes will give us non prime results, so:
[tex]P(B and C)=\frac{1}{3}[/tex]
h) The probability of the sum of the chips is even or prime. There are two outcomes where the sum of the chips is even or prime, {3,10} since 3+10=13 and {3,5} since 3+5=8 so:
[tex]P(A or C)=\frac{2}{3}[/tex]
i) The probability of the sum of the chips is prime given that the sum of the chips is odd. There are two possible results where the sum of the chips is odd {3,10} and {5,10} and only one of those results is even, {3,10}, so
[tex]P(C given B)=\frac{1}{2}[/tex]
j) The probability of the sum of the chips is prime given that the sum of the chips is even. There is only one possible even result: {3,5} but that result isn't prime, so
[tex]P(C given A)=0[/tex]
k) The probability of the sum of the chips is not odd. There is only one outcome where the sum of the chips is not odd (even), {3,5} so:
[tex]P(not B)=\frac{1}{3}[/tex]
l) The probability of the sum of the chips is not prime. There are two outcomes where the sum of the chips is not prime, {3,5} and {5,10} so:
[tex]P(not C)=\frac{2}{3}[/tex]
Are events A and B mutually exclusive?
Yes, events A and B are mutually exclusive.
Why or why not?
Because the results can either be even or odd, not both.
Are events B and C mutually exclusive?
No, events B and C are not mutually exclusive.
Why or Why not?
Because the result can be both, odd and prime.
What is the next term of the geometric sequence? 3, -12, 48
Answer:
-192
Step-by-step explanation:
it is a geometric progression
r=-4
Entering 38.00 into the Price of Sneakers field Entering 6.00 into the Price field Entering 3.00 into the Price of Leather field True or False: You will no
Answer:
This question seems incorrect.
Kindly take a look again and re-state it properly to enable me give the most accurate answer.
Thank you
A binomial experiment consists of 11 trials. The probability of success on trial 4 is 0.41. What is the probability of success on trial 8?A. 0.71B. 0.41C. 0.39D. 0.84E. 0.14
Answer:
B. 0.41
Step-by-step explanation:
Binomial experiment:
In a binomial experiment, the probability of the success on each trial is always the same.
The probability of success on trial 4 is 0.41.
This means that the probability of success on trial 8, and all the other 10 trials, is of 0.41, and thus the correct answer is given by option B.
4,3,5,9,12,17,...what is the next number?
Answer:
The next number is going to be 21
Answer:
19
Step-by-step explanation:
4 even number
3,5,7 odd numbers
14 even
17, 19, 21 even
Trapezoid A B C D is shown. A diagonal is drawn from point B to point D. Sides B C and A D are parallel. Sides B A and C D are congruent. Angle C B D is 24 degrees and angle B A D is 116 degrees.
What is the measure of angle ABD in trapezoid ABCD?
24°
40°
64°
92°
Answer:
40 degrees un edge
Step-by-step explanation:
Answer:
The person above me got this correct, so the answer to this is 40! I just did the Unit Test and got a 100%!
(3b-4)(b+2) in standard form
Answer:
3b^2 + 2b -8
Step-by-step explanation:
* means multiply
^ means exponent
3b * b = 3b^2
3b * 2 = 6b
-4 * b = -4b
-4 * 2 = -8
3b^2 + 6b -4b -8
3b^2 + 2b -8
What is the solution set of the equation x2+3*-4=6
Answer:
x=9
Step-by-step explanation:
Eight more than one-half of a number is twenty-two. Find the number.
Answer:
Below.
Step-by-step explanation:
22-8 = 14x2 = 28.
Which function represents the graph below?
Answer:
The answer is the third one below
(a) The heights of male students in a college are thought to be normally distributed with mean 170 cm and standard deviation 7.
The heights of 5 male students from this college are measured and the sample mean was 174 cm.
Determine, at 5% level of significance, whether there is evidence that the mean height of the male students of this college is higher than 170 cm.
[6]
(b) (i) The result of a fitness trial is a random variable X which is normally distributed with mean μ and standard deviation 2.4 . A researcher uses the results from a random sample of 90 trials to calculate a
98% confidence interval for μ . What is the width of this interval?
[4]
(ii) Packets of fish food have weights that are distributed with standard deviation 2.3 g. A random sample of 200 packets is taken. The mean weight of this sample is found to be 99.2 g. Calculate a 99% confidence interval for the population mean weight.
[4]
(c) (i) Explain the difference between a point estimate and an interval
Estimate. [2]
(ii) The daily takings, $ x, for a shop were noted on 30 randomly chosen days. The takings are summarized by Σ x=31 500 and
Σ x2=33 141 816 .
Calculate unbiased estimates of the population mean and variance of the shop’s daily taking. [4
Answer:
the answer is 50 but I don't know if
What angles can you construct using just a pair of compasses and a ruler?
Answer:
By using a pair of compasses and a ruler you can draw all angles
Pls help me someone this is annoying me
Answer:
They are both 42 cm
Step-by-step explanation:
Integration of [(x+1)/(x-1)]dx
Hello!
∫[(x+1)/(x-1)dx
∫t+2/t dt
∫t/t + 2/t dt
∫1 + 2/t dt
∫1dt + ∫2/t dt
∫t + 2In (|t|)
x - 1 + 2In (|x-1|)
x + 2In (|x-1|) + C, C ∈ R
Good luck! :)
I pleased anyone to help me please
Answer:
The first one (90, 90) is supplimentary, the next two (54, 36. and 45, 45) are complimentary, and the last two are supplimentary.
Step-by-step explanation:
A complimentary angle is two angles that add up to 90, and supplimentary is two angles that add up to 180! :)
Answer:
1st picture at the top would be a supplementary angle because a supplementary angles always add to 180 degrees.
the 54 and 36 one is a complementary angle
the 45 and 45 would be complementary angle
the last two on the bottom would both be supplementary angles.
High hopes-
Barry
Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72
inches and standard deviation 3.17 inches. Compute the probability that a simple random sample of size n=
10 results in a sample mean greater than 40 inches. That is, compute P(mean >40).
Gestation period The length of human pregnancies is approximately normally distributed with mean u = 266
days and standard deviation o = 16 days.
Tagged
Math
1. What is the probability a randomly selected pregnancy lasts less than 260 days?
2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days
or less?
3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days
or less?
4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of
the mean?
Know
Learn
Booste
V See
Answer:
0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.
Gestation periods:
1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.
2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.
3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.
4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.
Step-by-step explanation:
To solve these questions, 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
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.
This means that [tex]\mu = 38.72, \sigma = 3.17[/tex]
Sample of 10:
This means that [tex]n = 10, s = \frac{3.17}{\sqrt{10}}[/tex]
Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.
This is 1 subtracted by the p-value of Z when X = 40. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}[/tex]
[tex]Z = 1.28[/tex]
[tex]Z = 1.28[/tex] has a p-value of 0.8997
1 - 0.8997 = 0.1003
0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.
Gestation periods:
[tex]\mu = 266, \sigma = 16[/tex]
1. What is the probability a randomly selected pregnancy lasts less than 260 days?
This is the p-value of Z when X = 260. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{260 - 266}{16}[/tex]
[tex]Z = -0.375[/tex]
[tex]Z = -0.375[/tex] has a p-value of 0.3539.
0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.
2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?
Now [tex]n = 20[/tex], so:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}[/tex]
[tex]Z = -1.68[/tex]
[tex]Z = -1.68[/tex] has a p-value of 0.0465.
0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.
3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?
Now [tex]n = 50[/tex], so:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}[/tex]
[tex]Z = -2.65[/tex]
[tex]Z = -2.65[/tex] has a p-value of 0.0040.
0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.
4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?
Sample of size 15 means that [tex]n = 15[/tex]. This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.
X = 276
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}[/tex]
[tex]Z = 2.42[/tex]
[tex]Z = 2.42[/tex] has a p-value of 0.9922.
X = 256
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}[/tex]
[tex]Z = -2.42[/tex]
[tex]Z = -2.42[/tex] has a p-value of 0.0078.
0.9922 - 0.0078 = 0.9844
0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.
HELPPPPPPP PLEASEEEEEEE
Answer:
150 dollars. if I am wrong correct me
Answer:
C and D
Step-by-step explanation:
15 to 30 galons at $9.95 to $21.00
the minimum amount can be found by calculating the minimum amount sold at a minimum price 15*9.95 = $149.25
the maximum amount can be found by calculating the maximum amount sold at a maximum price 30*21 = $630
there are 2 choices that are between 149.25 and 630, C, and D
What is the point estimate for the number of cars sold per week for a sample consisting of the following weeks: 1, 3, 5, 7, 10, 13, 14, 17, 19, 21?
A.
4.8
B.
5.22
C.
6.38
D.
6.1
Answer: A.
Step-by-step explanation:
Hope this helps!
1. S = 10 mm
V= S×S×S
=___×___×___
=____ mm3
Hi there!
»»————- ★ ————-««
I believe your answer is:
[tex]V=1000\text{mm}^3[/tex]
»»————- ★ ————-««
Here’s why:
⸻⸻⸻⸻
I am assuming by the infomation given that the figure is a cube.
⸻⸻⸻⸻
[tex]\boxed{\text{Finding the volume of the cube...}}\\\\S = 10mm; V= s^3\\--------------\\\rightarrow V = 10^3\\\\\rightarrow V = 10 * 10 * 10\\\\\rightarrow \boxed{V=1000\text{mm}^3}[/tex]
⸻⸻⸻⸻
»»————- ★ ————-««
Hope this helps you. I apologize if it’s incorrect.
Write a polynomial f (x) that satisfies the given conditions. Polynomial of lowest degree with zeros of -4 (multiplicity 3), 1 (multiplicity 1), and with f(0) = 320.
Answer:
Step-by-step explanation:
Polynomial f(x) has the following conditions: zeros of -4 (multiplicity 3), 1 (multiplicity 1), and with f(0) = 320.
The first part zeros of -4 means (x+4) and multiplicity 3 means (x+4)^3.
The second part zeros of 1 means (x-1) and multiplicity 1 means (x-1).
The third part f(0) = 320 means substituting x=0 into (x+4)^3*(x-1)*k =320
(0+4)^3*(0-1)*k = 320
-64k = 320
k = -5
Combining all three conditions, f(x)
= -5(x+4)^3*(x-1)
= -5(x^3 + 3*4*x^2 + 3*4*4*x + 4^3)(x-1)
= -5(x^4 + 12x^3 + 48x^2 + 64x - x^3 - 12x^2 - 48x - 64)
= -5(x^4 + 11x^3 + 36x^2 + 16x -64)
= -5x^3 -55x^3 - 180x^2 - 80x + 320
Answer:
Step-by-step explanation:
-4 is a root for 3 times and 1 is root for once
so (x+4)^3 * (x-1) is part of f(x)
the constant term there is 4^3*(-1)=-64
so there is a multiplier of 320/-64=-5
f(x) = -5 * (x+4)^3 * (x-1)
What is the range of the table of values
Answer:
Range: { 0,3,5,7,9}
Step-by-step explanation:
The range is the values that y takes
Range: { 0,3,5,7,9}
Now we have to find,
The range of the table of values,
→ Range = ?
Then the range will be the numbers that is in the Y column.
→ Range = ?
→ Range = (value that Y takes)
→ Range = 0,3,5,7,9
Therefore, the range is 0,3,5,7,9.
The distribution of the number of children for families in the United States has mean 0.9 and standard deviation 1.1. Suppose a television network selects a random sample of 1000 families in the United States for a survey on TV viewing habits.
Required:
a. Describe (as shape, center and spread) the sampling distribution of the possible values of the average number of children per family.
b. What average numbers of children are reasonably likely in the sample?
c. What is the probability that the average number of children per family in the sample will be 0.8 or less?
d. What is the probability that the average number of children per family in the sample will be between 0.8 and 1.0?
Answer:
a) By the Central Limit Theorem, it has an approximately normal shape, with mean(center) 0.9 and standard deviation(spread) 0.035.
b) Average numbers of children between 0.83 and 0.97 are reasonably likely in the sample.
c) 0.0021 = 0.21% probability that the average number of children per family in the sample will be 0.8 or less
d) 0.9958 = 99.58% probability that the average number of children per family in the sample will be between 0.8 and 1.0
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 0.9 and standard deviation 1.1.
This means that [tex]\mu = 0.9, \sigma = 1.1[/tex]
Suppose a television network selects a random sample of 1000 families in the United States for a survey on TV viewing habits.
This means that [tex]n = 1000, s = \frac{1.1}{\sqrt{1000}} = 0.035[/tex]
a. Describe (as shape, center and spread) the sampling distribution of the possible values of the average number of children per family.
By the Central Limit Theorem, it has an approximately normal shape, with mean(center) 0.9 and standard deviation(spread) 0.035.
b. What average numbers of children are reasonably likely in the sample?
By the Empirical Rule, 95% of the sample is within 2 standard deviations of the mean, so:
0.9 - 2*0.035 = 0.83
0.9 + 2*0.035 = 0.97
Average numbers of children between 0.83 and 0.97 are reasonably likely in the sample.
c. What is the probability that the average number of children per family in the sample will be 0.8 or less?
This is the p-value of Z when X = 0.8. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.8 - 0.9}{0.035}[/tex]
[tex]Z = -2.86[/tex]
[tex]Z = -2.86[/tex] has a p-value of 0.0021
0.0021 = 0.21% probability that the average number of children per family in the sample will be 0.8 or less.
d. What is the probability that the average number of children per family in the sample will be between 0.8 and 1.0?
p-value of Z when X = 1 subtracted by the p-value of Z when X = 0.8.
X = 1
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1 - 0.9}{0.035}[/tex]
[tex]Z = 2.86[/tex]
[tex]Z = 2.86[/tex] has a p-value of 0.9979
X = 0.8
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.8 - 0.9}{0.035}[/tex]
[tex]Z = -2.86[/tex]
[tex]Z = -2.86[/tex] has a p-value of 0.0021
0.9979 - 0.0021 = 0.9958
0.9958 = 99.58% probability that the average number of children per family in the sample will be between 0.8 and 1.0
The linear equation Y = a + bX is often used to express cost formulas. In this equation:_________
a) the b term represents variable cost per unit of activity.
b) the a term represents variable cost in total.
c) the X term represents total cost.
d) the Y term represents total fixed cost.
find the measures of m and n.
Answer:
m = 4
n = 5
Step-by-step explanation:
[tex]m + 8 = 3m\\\\m - 3m = - 8\\\\-2m = - 8\\\\m = 4[/tex]
[tex]2n - 1 = 9 \\\\2n = 9 + 1\\\\2n = 10\\\\n = 5[/tex]
Find m angle JRQ if m angle SRQ=166^ and m angle SRJ=110^
Answer:
[tex] \large{ \tt{❃ \: S \: O \: L \: U \: T \: I \: O \: N : }}[/tex]
[tex] \large{ \tt{❉ \: m \: \angle \:SRQ = m \: \angle \: SRJ\: + \: m \: \angle \:JRQ}}[/tex]
[tex] \large{ \tt{⟼ \: 166 \degree = 110 \degree + m \: \angle \: JRQ}}[/tex]
[tex] \large{ \tt{⟼ \: 166 \degree - 110 \degree = m \: \angle \: JRQ}}[/tex]
[tex] \boxed{ \large{ \tt{⟼ \: 56 \degree = m \: \angle \: JRQ}}}[/tex]
Our final answer is 56° . Hope I helped! Let me know if you have any questions regarding my answer! :)Which best describes the function represented by the
table?
Х
-2
2
4
6
Y у
-5
5
10
15
O direct variation; k = 33 를
O direct variation; k = 5
- 를
O inverse variation; k = 10
direct variation; k = 1
10
Answer:
Direct variation
[tex]k = 2.5[/tex]
Step-by-step explanation:
Given
The attached table
Required
The type of variation
First, we check for direct variation using:
[tex]k = \frac{y}{x}[/tex]
Pick corresponding points on the table
[tex](x,y) = (-2,-5)[/tex]
So:
[tex]k = \frac{-5}{-2} = 2.5[/tex]
[tex](x,y) = (4,10)[/tex]
So:
[tex]k = \frac{10}{4} = 2.5[/tex]
[tex](x,y) = (6,15)[/tex]
So:
[tex]k = \frac{15}{6} = 2.5[/tex]
Hence, the table shows direct variation with [tex]k = 2.5[/tex]
how to solve for
LN and what are the variables
Answer:
v See below. v
Step-by-step explanation:
LM = MN
11x - 21 = 8x + 15
[tex]3x-21=15\\3x=36\\[/tex]
x = 12
LM = 11(12) - 21 = 132 - 21 = 111
MN = 8(12) + 15 = 96 + 15 = 111
LN = 111 + 111 = 222
5t/4y=3b/4c (solve for y)
I also need to know the steps.
thanks.
Answer:
[tex]y = \frac{5ct}{3b}[/tex]
Step-by-step explanation:
[tex]\frac{5t}{4y} =\frac{3b}{4c}[/tex]
1. start by multiplying y to both sides:
y × [tex]\frac{5t}{4y} =\frac{3b}{4c}[/tex] × y
[tex]\frac{5t}{4} =\frac{3b}{4c}y[/tex]
2. divide both sides by [tex]\frac{3b}{4c}[/tex]
[tex]\frac{5t}{4}/\frac{3b}{4c} =\frac{3b}{4c}y/\frac{3b}{4c}[/tex]
[tex]y = \frac{5ct}{3b}[/tex]
Please help. I'm stuck on this problem
Answer:
Step-by-step explanation:
[tex]h(t)=-16t^2+96t\\\\h(t)=-t(16t-96)[/tex]
[tex]96=2^5*3\\\\16=2^4\\\\h(t)=-t(2^5*3*t-2^4)=-2^4t(2^1*3*t-1)\\\\h(t)=-16t(6t-1)[/tex]
the b) part is easy do it!
is perpendicular to line segment
. If the length of is a units, then the length of is
units.
Answer:
AB is perpendicular to [GH] and GH is [A]
Step-by-step explanation:
