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daser333 [38]
4 years ago
6

a golfers score after playing on Friday was +2. His score saturdays round was -5. At the wnd of his round on Sunday he was at ev

en par, or 0. What integer relresents the change im the golfers score from the end of his round on saturday to end of his round on sunday?
Mathematics
2 answers:
Oksana_A [137]4 years ago
4 0

Score of golfer after playing on Friday= +2

Score of golfer on Saturday= -5

After Sunday round , Score of golfer =Either an Even Integer or 0.

The Integer which represents,the change  the golfers score from the end of his round on saturday to end of his round on sunday

                = Score of golfer on Sunday -  Score of golfer on Saturday

                  = 0 - (-5)    

                 = +5

or

An Even Integer - (-5)

=An Even Integer + (5)

=An Integer greater than 5

zmey [24]4 years ago
3 0

5. The end of his round on Saturday was a -5, and the end of his round on Sunday was a 0. The difference between -5 and 0 is 5, which means that the integer that represents the change in the golfers score is 5.

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4 years ago
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3 years ago
A student carried out an experiment to determine the amount of vitamin C in a tablet sample. He performed 5 trials to produce th
ivolga24 [154]

Answer:

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P-value = 0.166.

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{5}(490+502+505+495+492)\\\\\\M=\dfrac{2484}{5}\\\\\\M=496.8\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}((490-496.8)^2+(502-496.8)^2+(505-496.8)^2+(495-496.8)^2+(492-496.8)^2)}\\\\\\s=\sqrt{\dfrac{166.8}{4}}\\\\\\s=\sqrt{41.7}=6.5\\\\\\

Then, we can perform the hypothesis t-test for the mean.

The claim is that the amount of vitamin C in a tablet sample is different from 500 mg.

Then, the null and alternative hypothesis are:

H_0: \mu=500\\\\H_a:\mu< 500

The significance level is 0.05.

The sample has a size n=5.

The sample mean is M=496.8.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.5.

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

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6.5}{\sqrt{5}}=2.907

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{496.8-500}{2.907}=\dfrac{-3.2}{2.907}=-1.1

The degrees of freedom for this sample size are:

df=n-1=5-1=4

This test is a left-tailed test, with 4 degrees of freedom and t=-1.1, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t

As the P-value (0.166) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

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