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chubhunter [2.5K]

4 years ago

15

DO NOT POST UNUSEFUL ANSWERS! Serious answers only! A softball has a circumference of 12 inches. What is the softballs approxima

te volume? A. 16.41 in^3 B. 24.56 in^3 C. 29.18 in^3 or D. 31.27 in^3

2 answers:

abruzzese [7]4 years ago

5 0

Answer:

The answer would be 29.18 in³

Step-by-step explanation:

First we have to find the radius to figure out the volume of the softball.

Since we have the circumference of the softball, we would divide the total amount by pi, and then divide the answer by 2 (because radius is half of diameter). After this you would take the radius, raise it to the power of 3, and then multiply it by 4/3 and pi. This is because the volume of a sphere is:

V= $\frac{4}{3}$ πr³.

After substituting in everything, we would get our final answer of 29.18 in³.

(If you thought my answer was good, pls consider giving me a Brainliest!)

xxMikexx [17]4 years ago

4 0

Formula for volume of a sphere: V = (4 / 3)(pi)(r^3)

Formula for circumference: C = (pi)(2r)

Solving for radius using circumference formula:

12 = (pi)(2r)

2r = 12 / pi

r = 12 / 2pi

Solving for volume:

V = (4 / 3)(pi)((12 / (2pi))^3)

Plug into calculator

V = 29.18 in^3

Hope this helps!! :)

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90 hazelnuts have a mass of 125 g. Hazelnuts have 630 calories per 100 g. work out the number of calories per hazelnut

jenyasd209 [6]

90 hazelnuts = 125g

Therefore, the mass of one hazelnut is this...

90 / 90 = 125 / 90

1 hazelnut = 25/18 grams

630 calories = 100 grams

100 divided by 25/18 = 72

Therefore...

630 / 72 = 1 hazelnut

AND THE ANSWER IS

8.75 calories = 1 hazelnut

8 0

4 years ago

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lawyer [7]

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3 years ago

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11 chocolate bars cost 10 dollars

inna [77]

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Solution: Cost of 1 candy bar is $0.91

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-Austint1414

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3 years ago

Consider the computer output below. Fill in the missing information. Round your answers to two decimal places (e.g. 98.76). Test

slamgirl [31]

Answer:

$SE_{Mean}=\frac{s}{\sqrt{n}}=\frac{4.77}{\sqrt{19}}=1.094$

$t=\frac{98.77-100}{\frac{4.77}{\sqrt{19}}}=-1.124$

The 95% confidence interval would be given by (96.625;100.915)

a) $df=n-1= 19-1= 18$

b) $p_v =2*P(t_{18}$

If we compare the p value and a significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

c) The only thing that changes is the p value and would be given by:

$p_v =P(t_{18}>-1.124)=0.862$

But again since the p value is higher than the significance level we fail to reject the null hypothesis.

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

$\bar X=98.77$ represent the sample mean

$s=4.77$ represent the sample standard deviation

n=19 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if we have significant difference on the mean of 100, the system of hypothesis would be:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we can calculate the Standard error for the mean like this:

$SE_{Mean}=\frac{s}{\sqrt{n}}=\frac{4.77}{\sqrt{19}}=1.094$

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$98.77-1.96\frac{4.77}{\sqrt{19}}=96.625$

$98.77+1.96\frac{4.77}{\sqrt{19}}=100.915$

So on this case the 95% confidence interval would be given by (96.625;100.915)

Part a

The degree of freedom are given by:

$df=n-1= 19-1= 18$

Part b

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{98.77-100}{\frac{4.77}{\sqrt{19}}}=-1.124$

Then since is a two sided test the p value would be:

$p_v =2*P(t_{18}$

If we compare the p value and a significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Part c

If the system of hypothesis on this case are:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu > 100$

The only thing that changes is the p value and would be given by:

$p_v =P(t_{18}>-1.124)=0.862$

But again since the p value is higher than the significance level we fail to reject the null hypothesis.

7 0

3 years ago

The equation k = 1.6d gives an approximate relationship between d miles and k kilometers. The distance between Boise, Idaho, and

Burka [1]

Answer: C) 688 km

=================================================

Work Shown:

k = 1.6d

k = 1.6*430

k = 688

430 miles is roughly equivalent to 688 km.

More accurately, 430 miles = 692.01792 km, so this approximation isn't too far off.

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3 years ago

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