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KATRIN_1 [288]
3 years ago
7

Please help me with this one. N° 15

Mathematics
1 answer:
Ivenika [448]3 years ago
8 0
60° because its not hafe evin and 90° is hafe so its 60°
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Need help at this question
olga2289 [7]
The answer is definitely 2.4>3x
7 0
3 years ago
Jason has three pairs of pants, five shirts, and two pairs of shoes from which to choose to wear to school. How many possible ou
irina1246 [14]

Answer:

30

Step-by-step explanation:

3 pants * 5 shirts * 2 shoes

30 out fits

4 0
3 years ago
Nike claims that the number of miles a jogger can get a on a pair of Nike’s running shoes is higher than 1000. Moreover, Nike al
Lisa [10]

Answer:

a) There is statistical evidence showing that Nike shoues get more than 1000 miles.

b) The p-value of t=1.84 and df=149 is P=0.034.

It represents the probability of getting this sample results if the population parameters are the ones from the null hypothesis.

In this case, the low p-value shows is rare to get a sample of n=150 and mean 1015 miles, if the population mean is 1000. This is evidence that the null hypothesis is not true.

c) There is no enough evidence to claim that Nike shoes outperform Adidas shoes by 15 miles.

d) The p-value for t=0.55 and 318 df is P=0.29.

This shows that there is a probability of P=0.29 of getting samples that show this difference if the statement of the null hypothesis is true.

Step-by-step explanation:

a) We have to perform a hypothesis test with the following hypothesis:

H_0: \mu\leq 1000\\\\H_a: \mu>1000

The level of significance is 0.05.

The sample mean is 1015 and the sample standard deviation is 100.

The degrees of freedom are df = 150-1 = 149.

The t-statistic is:

t=\frac{M-\mu}{s/\sqrt{n}}=\frac{1015-1000}{100/\sqrt{150}} =\frac{15}{8.165} =1.84

The critical value for t is t=1.66. As the t-statistic is bigger than the critical t, it falls in the rejection region. The null hypothesis is rejected.

There is statistical evidence showing that Nike shoues get more than 1000 miles.

b) The p-value of t=1.84 and df=149 is P=0.034.

It represents the probability of getting this sample results if the population parameters are the ones from the null hypothesis.

In this case, the low p-value shows is rare to get a sample of n=150 and mean 1015 miles, if the population mean is 1000. This is evidence that the null hypothesis is not true.

c) In this case, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2\leq 15\\\\H_a: \mu_1-\mu_2>15

The significance level is 0.05.

The difference between sample means is:

M_d=1015-995=20

The standard deviation of the difference is:

\sigma=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}= \sqrt{\frac{100^2}{150}+\frac{50^2}{170}}= \sqrt{66.67+14.70}=9

The degrees of freedom are:

df=n_1+n_2-2=150+170-2=318

The critical value is t=1.65

The t-statistic is:

t=\frac{M_d-(\mu_1-\mu_2)}{\sigma} =\frac{20-15}{9}= \frac{5}{9}= 0.55

The value ot the t-statistic is lower than the critical value, so it lies within the acceptance region.

There is no enough evidence to claim that Nike shoes outperform Adidas shoes by 15 miles.

d) The p-value for t=0.55 and 318 df is P=0.29.

This shows that there is a probability of P=0.29 of getting samples that show this difference if the statement of the null hypothesis is true.

6 0
3 years ago
LL Incorporated’s currently outstanding 7 percent coupon bonds have a yield to maturity of 6 percent. LL believes it could sell
Ipatiy [6.2K]

Answer:

4.2%

Step-by-step explanation:

Data provided in the question:

yield to maturity of coupon bonds = 6 percent = 0.06

marginal tax rate = 30 percent = 0.3

Now,

The LL’s after-tax cost of debt is calculated using the formula

= yield to maturity × ( 1 - marginal tax rate )

on substituting the respective values, we get

= 0.06 × ( 1 - 0.30 )

or

= 0.06 × 0.7

or

= 0.042

or

= 0.042 × 100%

= 4.2%

6 0
3 years ago
Jerry currently has an account of $1,109.80. His initial deposit on the account was $775 and it earned 2.7% simple interest. How
Gekata [30.6K]
I=prt
1109.80=(775)(0.027)(t)
t=53 years
6 0
3 years ago
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