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3 years ago

7y-9=3y+19. Need help

Mathematics

2 answers:

CaHeK987 [17]3 years ago

8 0

Answer: y = 7.

Step-by-step explanation:

First you need to combine like terms. The like terms in this is 7y & 3y and -9 & 19. So to combine those like terms, you first need to subtract 3y on both sides because to cancel out the 3y in the right side, you need to perform the opposite operation on the positive 3. And 3y - 3y is 0 so thats why you subtract 3y on both sides. When you subtract 3y from 7y on the left side (because 3y is the like term of 7y and what operation you do on one side you have to do on the other side of the equation) you get 4y - 9 = 19. Now you still need to combine the like terms -9 and 19 so you need to add 9 on both sides to get 4y = 28. Since 4 is being multiplied by "y" to get "y" by itself in the left side of the equation, you need to divide 4 on both sides. 28 ÷ 4 = 7, so your final answer would be y = 7!

I hope this is helpful :)

alexgriva [62]3 years ago

3 0

Answer:

y=7

Step-by-step explanation:

7y-3y=19+9

4y=28

y=7

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Exercise 11.21) Many smartphones, especially those of the LTE-enabled persuasion, have earned a bad rap for exceptionally bad ba

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Answer:

Step-by-step explanation:

Hello!

The researcher suspects that the battery life between charges for the Motorola Droid Razr Max differs if its primary use is talking or if its primary use is for internet applications.

Since the means for talk time usage (20hs) is greater than the mean for internet usage (7hs) the main question is if the variance in hours of usage is also greater when the primary use is talk time.

Be:

X₁: Battery duration between charges when the primary usage of the phone is talking. (hs)

n₁= 12

X[bar]₁= 20.50 hs

S₁²= 199.76hs² (S₁= 14.13hs)

X₂: Battery duration between charges when the primary usage of the phone is internet applications.

n₂= 10

X[bar]₂= 8.50

S₂²= 33.29hs² (S₂= 5.77hs)

Assuming that both variables have a normal distribution X₁~N(μ₁;σ₁²) and X₂~N(μ₂; σ₂²)

The parameters of interest are σ₁² and σ₂²

a) They want to test if the population variance of the duration time of the battery when the primary usage is for talking is greater than the population variance of the duration time of the battery when the primary usage is for internet applications. Symbolically: σ₁² > σ₂² or since the test to do is a variance ratio: σ₁²/σ₂² > 1

The hypotheses are:

H₀: σ₁²/σ₂² ≤ 1

H₁: σ₁²/σ₂² > 1

There is no level of significance listed so I've chosen α: 0.05

b) I've already calculated the sample standard deviations using a software, just in case I'll show you how to calculate them by hand:

S²= $\frac{1}{n-1}$ *[∑X²-(∑X)²/n]

For the first sample:

n₁= 12; ∑X₁= 246; ∑X₁²= 7240.36

S₁²= $\frac{1}{11}$ *[7240.36-(246)²/12]= 199.76hs²

S₁=√S₁²=√199.76= 14.1336 ≅ 14.13hs

For the second sample:

n₂= 10; ∑X₂= 85; ∑X₂²= 1022.12

S₂²= $\frac{1}{9}$ *[1022.12-(85)²/10]= 33.2911hs²

S₂=√S₂²=√33.2911= 5.7698 ≅ 5.77hs

For this hypothesis test, the statistic to use is a Snedecors F:

$F= \frac{S_1^2}{S_2^2} *\frac{Sigma_1^2}{Sigma_2^2} ~~F_{n_1-1;n_2-1}$

This test is one-tailed right, wich means that you'll reject the null hypothesis to big values of F:

$F_{n_1-1;n_2-1; 1-\alpha }= F_{11;9;0.95}= 3.10$

The rejection region is then F ≥ 3.10

$F_{H_0}= \frac{199.76}{33.29} * 1= 6.0006$

p-value: 0.006

Considering that the p-value is less than the level of significance, the decision is to reject the null hypothesis.

Then at a 5% level, there is significant evidence to conclude that the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for talk is greater than the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for internet applications.

I hope this helps!