FIND THE SURFACE AREA PLEASE

e-lub [12.9K]

B = rising slowly
C = constant
D = falling quickly

This graph could be an example of the Microsoft Share in the stock market.

3 0

4 years ago

A random sample of 25 fields of spring wheat has a mean yield of 27.6 bushels per acre and standard deviation of 5.69 bushels pe

ale4655 [162]

Answer:

Step 1 z(c) = 2,323

CI = ( 25 ; 30,2 )

Step-by-step explanation:

Normal Distribution:

Sample size n = 25

Sample mean μ = 27,6

Sample standard deviation s = 5,69

CI = 98 % then α = 1 - 0,98 α = 0,02 α/2 = 0,01

From z-table we don´t find z (score) for 0,01 directly, we need to interpolate between z = 2,32 and z = 2,33

For 0,0099 z score is 2,33

for 0,0102 z score is 2,32

Δ 0,0003 0,01

Then by rule of three

for Δ 0,0003 ⇒ 0,01

for Δ (0,01- 0,0102) ⇒ x

0,0003 0,01

0,0002 x

x = 0,0067

then z (score for 0,01) = 2,33 - 0,0067 z(c) = 2,3233

round to three decimal places z(c) = 2,323

Step 2:

CI = μ ± z(c) * s/√n ⇒ 27,6 ± (2,323 * 5,69)/5

CI = 27,6 ± 2,6436

round to one decimal place

CI = 27,6 ± 2,6

CI = ( 25 ; 30,2 )

8 0

3 years ago

Which has one line of symmetry

Darya [45]

Answer:

D

Step-by-step explanation:

An isoceles triangle has two eqial sides and angles wich give it exactly one axe of symmetry

8 0

3 years ago

Read 2 more answers

Solve. 5x – 10 ≤ 20 <br> A. (–∞, 2] <br> B. (–∞, 2)<br> C. (–∞, 6]<br> D. (–∞, 6)

elena55 [62]

The answer is C.

I hope this helps.

4 0

4 years ago

Read 2 more answers

A data set lists earthquake depths. The summary statistics arenequals=500500,x overbarxequals=5.435.43km,sequals=4.864.86km. U

Eva8 [605]

Answer:

C. H 0:μ=5.00km H1:μ≠5.00km

$z=\frac{5.43-5.00}{\frac{4.86}{\sqrt{500}}}=1.978$

$p_v =2*P(z>1.978)=0.048$

If we compare the p value and the significance level given $\alpha=0.01$ 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 true mean is nto significantly different from 5.00.

Step-by-step explanation:

Some previous concepts

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct.

A z-test for one mean "is a hypothesis test that attempts to make a claim about the population mean(μ)".

The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"

The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"

Data given and notation

$\bar X=5.43$ represent the average for the sample

$s=4.86$ represent the sample standard deviation for the sample

$n=500$ sample size

$\mu_o =5.00$ represent the value that we want to test

$\alpha=0.01$ 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 population mean is equal to 5.00, the system of hypothesis would be:

Null hypothesis: $\mu = 5.00$

Alternative hypothesis: $\mu \neq 5.00$

Since the sample size >30 and large, the estimator for the population deviation would be s, and we can use the 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:

$z=\frac{5.43-5.00}{\frac{4.86}{\sqrt{500}}}=1.978$

P-value

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

$p_v =2*P(z>1.978)=0.048$

In Excel we can us ethe following formula to find the p value "=2*(1-NORM.DIST(1.978;0;1;TRUE))"

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ 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 true mean is nto significantly different from 5.00.

8 0

3 years ago