PLEASE HELP WITH THIS MATH QUESTION I am so frustrated I can't get it right and this is my last test attempt :(

I had $88. then i earned $29 walking dogs. after I spent money for concert tickets, I have $74 remaining. estimate how much mone

klemol [59]

Answer:

$43

Step-by-step explanation:

88+29=117

117-74=43

You Spent $43 For The Concert.

7 0

4 years ago

Read 2 more answers

Suppose a professional golfing association requires that the standard deviation of the diameter of a golf ball be less than 0.00

Bingel [31]

Answer:

We conclude that the standard deviation of the diameter of a golf ball is less than 0.005 inch.

Step-by-step explanation:

We are given that a professional golfing association requires that the standard deviation of the diameter of a golf ball be less than 0.005 inch.

Assume that the population is normally distributed: 1.678, 1.681, 1.676, 1.684, 1.676, 1.679, 1.681, 1.681, 1.677, 1.676, 1.681, 1.683.

Let $\sigma$ = population standard deviation of the diameter of a golf ball.

SO, Null Hypothesis, $H_0$ : $\sigma \geq$ 0.005 inch {means that the standard deviation of the diameter of a golf ball is more than or equal to 0.005 inch}

Alternate Hypothesis, $H_0$ : $\sigma$ < 0.005 inch {means that the standard deviation of the diameter of a golf ball is less than 0.005 inch}

The test statistics that would be used here One-sample Chi-square test statistics;

T.S. = $\frac{(n-1)s^{2} }{{\sigma^{2} }}$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.00281 inch

n = sample size = 12

So, the test statistics = $\frac{(12-1)\times 0.00281^{2} }{{0.005^{2} }}$ ~ $\chi^{2} __1_1$

= 3.47

The value of chi-square test statistics is 3.47.

Also, the P-value of test statistics is given by the following formula;

P-value = P( $\chi^{2} __1_1$ < 3.47) = 0.0182

Since, the P-value of the test statistics is less than the level of significance as 0.0182 < 0.05, so we reject our null hypothesis.

Now, at 0.05 significance level the chi-square table gives critical value of 4.575 at 11 degree of freedom for left-tailed test.

Since our test statistic is less than the critical value of chi-square as 3.47 < 4.575, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the standard deviation of the diameter of a golf ball is less than 0.005 inch.

7 0

3 years ago

Estimate the area of the rectangle. A)8 B)10 C)14 D)20

seraphim [82]

Answer:

A or B, I think it's A...

7 0

3 years ago

14. A $40,000 loan at 4% dated June 10 is due to be paid on October 11. Calculate the amount of interest (assume ordinary intere

Allisa [31]

D) $546.67

Hope this helped!