I need help with this tysm

zloy xaker [14]

A is a square matrix

8 0

2 years ago

The results of a poll show that the percent of people who want a new restaurant is in the interval (62%, 88%)(62%, 88%) . There

xenn [34]

Required percentage = (0.62 x 109482, 0.88 x 109,482) = (67879, 96,344)

5 0

3 years ago

Read 2 more answers

What is the square root of 1/25

wolverine [178]

First simplify the square root of 1/25

-√1/25

Rewrite using the quote by rule for radicals.

-√1/√25

Any root of 1 is 1

-1/√25

Simplify the denominator. (find the square root of 25)

-1/5

The result can be shown in both exact and decimal forms.
Exact form: -1/5
Decimal form: -0.2

6 0

3 years ago

If<br> DX/DT= 2cm/s <br> x=-1 <br> y=x^2+1<br> DY/DT=??

kobusy [5.1K]

Answer:

$\frac{dy}{dt}$ = - 4 cm/s

Step-by-step explanation:

Let us revise the chain rule

If $\frac{dy}{dt}$ = a and $\frac{dx}{dt}$ = b, then

$\frac{dy}{dx}$ = $\frac{dy}{dt}$ ÷ $\frac{dx}{dt}$ = $\frac{a}{b}$

∵ y = x² + 1

- Use the differentiation to find $\frac{dy}{dx}$

∴ $\frac{dy}{dx}$ = 2x

∵ x = -1

- Substitute x by -1 in $\frac{dy}{dx}$

∴ $\frac{dy}{dx}$ = 2(-1)

∴ $\frac{dy}{dx}$ = -2

∵ $\frac{dy}{dt}$ = $\frac{dy}{dx}$ × $\frac{dx}{dt}$

∵ $\frac{dx}{dt}$ = 2 cm/s

∴ $\frac{dy}{dt}$ = (-2) × (2)

∴ $\frac{dy}{dt}$ = - 4 cm/s

8 0

3 years ago

Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 4.6 parts

kirill115 [55]

Answer:

$z=\frac{4.5-4.6}{\frac{1.1}{\sqrt{860}}}=-2.666$

$p_v =2*P(Z$

If we compare the p value and the significance level given $\alpha=0.02$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 4.6 at 2% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=4.5$ represent the sample mean

$\sigma=1.1$ represent the population standard deviation

$n=860$ sample size

$\mu_o =4.6$ represent the value that we want to test

$\alpha=0.02$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal or not to 4.6, the system of hypothesis would be:

Null hypothesis: $\mu = 4.6$

Alternative hypothesis: $\mu \neq 4.6$

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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".

Calculate the statistic

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

$z=\frac{4.5-4.6}{\frac{1.1}{\sqrt{860}}}=-2.666$

P-value

Since is a two sided test the p value would be:

$p_v =2*P(Z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.02$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 4.6 at 2% of signficance.

5 0

3 years ago

I need help with this tysm​

I need help with this tysm