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3 years ago

Pleaseeee help!!!!! I will mark you as brainlinest for correct answer!!!!!!

Mathematics

1 answer:

Setler79 [48]3 years ago

4 0

So the equation you'll use is (18,000 × .041)y+18000. Y is your years which is 20. So all you need to do is plug it in to get (18,000 × .041)20+18000 and when solved is 32,760 downloads by 2030

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-26a-19-35=-84 . show work!!

Wewaii [24]

Step-by-step explanation:

-26a-19-35=-84

-26a-19+(19)-35=-84+(19)

-26a-35=-65

-26a-35+(35)=-65+(35)

-26a=-30

-26a/26=-30/26

7 0

3 years ago

Evaluate the integral. (remember to use absolute values where appropriate. Use c for the constant of integration.) 5 cot5(θ) sin

ozzi

$I=5\int \frac{cos^{4}\theta }{sin\theta }\times cos\theta d\theta \\\\I=5\int \left ( 1-sin^{2}\theta \right )^{2}\times \frac{cos\theta }{sin\theta }d\theta \\put\ \sin\theta =t\\\\dt=cos\theta d\theta \\\\I=5\int\frac{t^{4}+1-2t^{2}}{t}dt\ \ \ \ \ \ \ \ \ \ \because (a-b)^2=a^2+b^2-2ab\\\\I=5\left ( \int t^{3}dt + \int \frac{1}{t} -2\int t \right )dt$

by using the integration formula

we get,

$\\I=5\left ( \frac{t^{4}}{4} +logt -t^{2}\right )\\\\I=\frac{5}{4}t^{4}+5\log t-5t^{2}+c$

now put the value of t=\sin\theta in the above equation

we get,

$\int 5\cot^5\theta \sin^4\theta d\theta=\frac{5}{4}sin^{4}\theta+5\log \sin\theta - 5sin^{2} \theta+c$

hence proved

7 0

3 years ago

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

adelina 88 [10]

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".

5 0

3 years ago

Foure ABCD is a parallelogram<br> What is the value of x<br><br><br> please help

Zigmanuir [339]

Answer:

x = 7

Step-by-step explanation:

The property of a parallelogram is the opposite sides are congruent, hence

AD = BC ← substitute values

5x + 3 = 38 ( subtract 3 from both sides )

5x = 35 ( divide both sides by 5 )

x = 7

6 0

3 years ago

Suppose 45% of the population has a college degree.

levacccp [35]

Using the normal distribution, there is a 0.2076 = 20.76% probability that the proportion of persons with a college degree will differ from the population proportion by greater than 3%.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as $np \geq 10$ and $n(1 - p) \geq 10$ .

The proportion estimate and the sample size are given as follows:

p = 0.45, n = 437.

Hence the mean and the standard error are:

$\mu = p = 0.45$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.45(0.55)}{437}} = 0.0238$

The probability that the proportion of persons with a college degree will differ from the population proportion by greater than 3% is <u>2 multiplied by the p-value of Z when X = 0.45 - 0.03 = 0.42</u>.

Hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem:

$Z = \frac{X - \mu}{s}$

Z = (0.42 - 0.45)/0.0238

Z = -1.26

Z = -1.26 has a p-value of 0.1038.

2 x 0.1038 = 0.2076.

0.2076 = 20.76% probability that the proportion of persons with a college degree will differ from the population proportion by greater than 3%.

More can be learned about the normal distribution at brainly.com/question/28159597

#SPJ1

8 0

2 years ago