A model satellite has a scale of 1 cm : 3 m. If the real satellite is 18 m wide, then how wide is the model satellite?

Alekssandra [29.7K]

Your insight that the function values differ by 5 every time is very important. So your f(x) will contain 5x as a term. You have to find a constant to add or subtract to make sure f(1)=-2.

f(x) = 5x-7 does the job.

7 0

3 years ago

Find the missing character (?)

4vir4ik [10]

The answer is 7, your welcome :).

6 0

4 years ago

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

Make V the subject of the formula E=mv2/2

Dima020 [189]

Answer:

v = ± $\sqrt{\frac{2E}{m} }$