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Helga [31]
3 years ago
14

Bilal's fishing line was on a spool with a radius of 4cm. Suddenly, a fish pulled on the line, and the spool spun 16 times befor

e Bilal began to reel in the fish. What is the distance the fish pulled the fishing line?
Mathematics
2 answers:
Aleks [24]3 years ago
8 0

First, I'd like to say that this question is flawed because the diameter of the spool changes as you pull the line out. Some would argue it's negligible I suppose.

At any rate, assuming there's a magic spool where the diameter doesn't change, let's find the cicumference so we can find the length of one wrap around the spool.

circumference = 2*pi*r = 2 * pi * 4cm = about 25.133 cm

Now if it turns 16 times we'll have 16 times the circumference.

16 * (25.133 cm)

= 402.128 cm

Alchen [17]3 years ago
5 0

Answer:

the answer is 402 because on khan it asks to round to the nearest cm

Step-by-step explanation:

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MAVERICK [17]

Answer:

a) $40920

b) $38500

Step-by-step explanation:

Given:

5 employees,

Mean = $40300

Median = $38500

Min = $32000

If he lowest paid employee gets a $3100 raise, then his salary becomes

$32000+$3100=$35100

a) If the mean was $40300, then the sum of 5 salaries is

\$40300\cdot 5=\$201500

After raising the lowest salary the sum becomes

\$201500+\$3100=\$204600

and new mean is

\dfrac{\$204600}{5}=\$40920

b) The  lowest salary becomes $35100. It is still smaller than the median, so the new median is the same as the old one.

New median = $38500

5 0
3 years ago
A random sample of n = 45 observations from a quantitative population produced a mean x = 2.5 and a standard deviation s = 0.26.
oee [108]

Answer:

P-value (t=2.58) = 0.0066.

Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean μ exceeds 2.4.

Then, the null and alternative hypothesis are:

H_0: \mu=2.4\\\\H_a:\mu> 2.4

The significance level is 0.05.

The sample has a size n=45.

The sample mean is M=2.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.26}{\sqrt{45}}=0.0388

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.5-2.4}{0.0388}=\dfrac{0.1}{0.0388}=2.58

The degrees of freedom for this sample size are:

df=n-1=45-1=44

This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>2.5801)=0.0066

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

7 0
3 years ago
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Answer:

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Step-by-step explanation:

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