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2 years ago

A company that manufactures golf balls produces a

Mathematics

2 answers:

liraira [26]2 years ago

4 0

Answer:

No, because all of the new type of balls are not hit first, the distances they travel cannot be compared to the distances of the original ball.

Mekhanik [1.2K]2 years ago

4 0

Answer:

Repeating the experiment will allow the company to compare the distances traveled by the new type of golf ball from both experiments.

Step-by-step explanation:

However, this is not exactly the best way to find out if the new generation of golf balls would travel further. A golf pro can be inconsistent in the way they hit each ball [room of error].

A better way to compare is to use a golf ball machine that would launch each ball at the same angle and with the same force. However, generally with more tries the room of error would decrease significantly.

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How can you simplify algebraic expressions

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Algebraic expressions are simplified by using given operators

like as arithmetic

But in algebraic expression there are terms

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2 years ago

What is the GCF of 54 and 32?

Musya8 [376]

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32: 2 2 2 2 2
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3 years ago

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Let f(x) = 2x3 – 4x.<br> Find f(-2).

N76 [4]

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Step-by-step explanation:

f(x) = 2x^3 - 4x.....find f(-2).....so sub in -2 for x

f(-2) = 2(-2^3) - 4(-2)

f(-2) = 2(-8) + 8

f(-2) = -16 + 8

f(-2) = -8 <==

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3 years ago

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Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

3 years ago

Alan bought 20 yards of fencing. He used 5.9 yards in one garden and 10.3 in another. Write remaining as a fraction in simplest

Over [174]

Its 3.8 because if you add 5.9 and 10.3 is 16.2, and if you subtract it by 20, it should equal 3.8.

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3 years ago

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