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2 years ago

The length of a rectangle is 3 less than twice the width. Write a function rule for the perimeter in terms of w, the width

Mathematics

2 answers:

Vlad [161]2 years ago

4 0

let's translate this word problem into math

the length (L) is (=) 3 less than (-3) twice the width (2*W).

L=-3+2W

we want to find the perimiter

the perimiter, P is P=2(L+W)

if we subsitute -3+2W for L from our first equation into our perimiter equation, we get

P=2(-3+2W+W)

P=2(-3+3W)

P=-6+6W

P=6W-6

so P(W)=6W-6 is the function of the perimiter in terms of the width

Nitella [24]2 years ago

3 0

l = 2w-3

P = 2(l+w)

substitute in for l

P = 2(2w-3 +w)

combine like terms

P = 2(3w-3)

P = 6w-6

P(w) = 6(w-1)

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A cell phone manufacturer claims that the average battery life of its newest flagship smartphone is exactly 20 hours. Javier bel

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Answer:

(B) Fail to reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours.

(C) There is not enough evidence at the α=0.05 level of significance to suggest that the true population mean battery life of the smartphone is less than 20 hours.

Step-by-step explanation:

1) Data given and notation

$\bar X=19.5$ represent the battery life sample mean

$s=1.9$ represent the sample standard deviation

$n=33$ sample size

$\mu_o =20$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean battery life is less than 20 :

Null hypothesis: $\mu \geq 20$

Alternative hypothesis: $\mu < 20$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{19.5-20}{\frac{1.9}{\sqrt{33}}}=-1.51$

4) P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=33-1=32$

Since is a one-side lower test the p value would be:

$p_v =P(t_{(32)}$

5) Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the average battery life it's not significantly different less than 20 hours at 5% of signficance. If we analyze the options given we have:

(A) Reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours. FALSE, we FAIL to reject the null hypothesis.

(B) Fail to reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours. TRUE, we fail to reject the null hypothesis that the mean would be 20 or higher .

(C) There is not enough evidence at the α=0.05 level of significance to suggest that the true population mean battery life of the smartphone is less than 20 hours. TRUE, we FAIL to reject the null hypothesis that the mean is greater or equal to 20 hours, so we reject the alternative hypothesis that the mean is less than 20 hours.

(D) There is enough evidence at the α=0.05 level of significance to support the claim that the true population mean battery life of the smartphone is not equal to 20 hours. FALSE the claim is not that the mean is different from 20. The real claim is: "Javier believes the mean battery life is less than 20 hours".

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3 years ago

What is the area of a circle with a radius of 7 cm? (Use 3.14 for , and round to the nearest tenth.)

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so the right answer is 153.9 cm ^2 