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3 years ago

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Mathematics

2 answers:

densk [106]3 years ago

8 0

Answer:

B. 2106 pi mm³

Step-by-step explanation:

the solution is attached,

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Marta_Voda [28]3 years ago

4 0

Answer:

B.2106π mm^3

Step-by-step explanation:

here is the volume of a drill

$\pi \: {r}^{2} h \: + \frac{1}{3} \pi \: {r}^{2} h$

$\pi \times 81 \times 18 + \frac{1}{3} \times \pi \times 81 \times 24$

$> > > 2106 \pi \: {mm}^{3}$

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The standard ratio of a photo's width to it's length is 2/3. what is the length of a photo that has a width of 14 inches?

ankoles [38]

L: w

Equation

2/3 = 14/ x

Cross multiply

2xX = 14x3

2x = 42

Eliminate

2x/2 = 42/2

X = 21

5 0

3 years ago

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Need help with question 13B i can not figure it out please help.

Vsevolod [243]

Short Answer 10
The tough part is the number of drinks.
Let the number of them be x
4.5 x + 3 + (10/100) * (4.5x + 3) + (7/100) * (4.5 x + 3) ≤ 60

The tip is not included in the tax, and the tax does not include the tip. You can shorten this up quite a bit.
4.5 x + 3 + (10/100) * (4.5x + 3) + (7/100) * (4.5 x + 3) ≤ 60
(1 + 10/100 + 7 /100) (4.5x + 3) ≤ 60 Combine the tax, tip, scone and Lattes

(1.17) * (4.5x + 3) ≤ 60 Divide by 1.17
4.5x + 3 ≤ 60 / 1.17
4.5x + 3 ≤ 51.28 Subtract 3
4.5x ≤ 48.28 Divide by 4.5
x ≤ 48.28 / 4.5
x ≤ 10.72

Discussion.
There's a problem. He can't buy 0.72 of a Latte. He can only buy 10 of them. In this case we must round down so what happens if he buys 10 lattes? Does the price the taxes and the tip allow him to get another Latte? The answer should be no because then you would be 11 lattes and the tip and taxes would push you over 60 dollars. But be careful, sometimes things like this can alter you answer..

Answer: 10 <<<<

7 0

3 years ago

With double-digit annual percentage increases in the cost of health insurance, more and more workers are likely to lack health i

Dahasolnce [82]

Answer:

1. χ² = 15.3902

2. The p value is :_____.a. less than .005

3. Conclusion

a. health insurance coverage is not independent of the size of the company

4. The percentages of employees

Small %= 33/50= 0.66

Medium %= 68/75 = 0.91

Large %= 88/100= 0.88

Step-by-step explanation:

1) We set up our null and alternative hypothesis as

H0: the employee health insurance coverage is independent of the size of the company

against the claim

Ha: the employee health insurance coverage is not independent of the size of the company

2) the significance level alpha is set at 0.05

3) the test statistic under H0 is

χ²= ∑ (O - E)²/ E where O is the observed and E is the expected frequency

which has an approximate chi square distribution with ( 3-1) (2-1)= 2 d.f

4) Computations:

Under H0 , the observed frequencies are :

Observed Expected E (O-E) (O-E)² (O-E)²/E

33 42 -9 81 1.928

68 63 5 25 0.3968

88 84 4 16 0.1904

17 8 9 81 10.125

7 12 -5 21 1.75

12 16 -4 16 1

15.3902

Expected Values are calculated using the formula :

Row Total * Column Total / sample size

E1= (33+17) (33+ 68+88)/ 50+75+100= 42

E4= (33+17) (17+ 7+ 12)/ 50+75+100=8

E2= (68+7) (33+ 68+88)/ 50+75+100= 63

E5= (68+7) (17+ 7+ 12)/ 50+75+100= 12

E3= (88+12) (33+ 68+88)/ 50+75+100= 84

E6= (88+12) (17+ 7+ 12)/ 50+75+100= 16

5) The critical region is χ² ≥ χ² (0.05)2 = 5.99

6) Conclusion:

The calculated χ² = 15.3902 falls in the critical region χ² ≥ 5.99 so we reject the null hypothesis that the employee health insurance coverage is NOT independent of the size of the company.

2. The p value is :_____.

a. less than .005

The p-value is .000385.

3. Conclusion

a. health insurance coverage is not independent of the size of the company

4. The percentages of employees

Small %= 33/50= 0.66

Medium %= 68/75 = 0.91

Large %= 88/100= 0.88

3 0

3 years ago

A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is current

lapo4ka [179]

Using the z-distribution, we have that:

The hypothesis test is:

$H_0: \mu = 7.4$

$H_1: \mu < 7.4$

The p-value is of 0.0668.

The critical value is $z^{\ast} = -1.28$

b) Since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the null hypothesis will be rejected.

Item a:

At the null hypothesis, it is tested if the mean is of 7.4 minutes, that is:

$H_0: \mu = 7.4$

At the alternative hypothesis, it is tested if the mean is lower than 7.4 minutes, that is:

$H_1: \mu < 7.4$

We have the standard deviation for the population, hence, the z-distribution is used.

The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.

$\mu$ is the value tested at the null hypothesis.

$\sigma$ is the standard deviation of the sample.

n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 7.1, \mu = 7.4, \sigma = 1.2, n = 36$ .

Hence, the value of the test statistic is:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{7.1 - 7.4}{\frac{1.2}{\sqrt{36}}}$

$z = -1.5$

Using a z-distribution calculator, the p-value is of 0.0668.

Also using a calculator, considering a left-tailed test, as we are testing if the mean is less than a value, the critical value with a significance level of 0.1 is of $z^{\ast} = -1.28$ .

Item b:

Since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the null hypothesis will be rejected.

You can learn more about the use of the z-distribution to test an hypothesis at brainly.com/question/16313918

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2 years ago

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Nookie1986 [14]

Answer:

Step-by-step explanation:

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6 0

3 years ago

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