All categories

3 years ago

Find the percentage decrease if a players batting average dropped from 0.273 to 0.265 in one season.

Mathematics

1 answer:

sesenic [268]3 years ago

4 0

Answer:

$\huge\boxed{2.93\%}$

Step-by-step explanation:

Calculate the difference:

$0.273-0.265=0.008$

Calculate the ratio:

$\dfrac{0.008}{0.273}=\dfrac{0.008\cdot1000}{0.273\cdot1000}=\dfrac{8}{273}$

Convert to percentage:

$\dfrac{8}{273}=\dfrac{8}{273}\cdot100\%=\dfrac{800}{273}\%\approx2.93\%$

You might be interested in

If the fourth multiple of 3 is subtracted from the third multiple of 4 what is the difference

anyanavicka [17]

The difference is zero if you aren't insinuating that the term, "multiple" doesn't involve exponential factors.

3•4 = 12
4•3 = 12

12
-12
___
0

However, if you are saying one multiple of 3 involves exponential powers, Then 3^4 is 81 and 4^3 is 64

4^3 = 64

64 - 81 = -17

8 0

2 years ago

Read 2 more answers

A killer whale weighs 4.6 tons. Convert the weight to pounds.

Lena [83]

Answer:

9200

Step-by-step explanation:

3 0

3 years ago

The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed

algol [13]

Answer:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

$p_v = P(\chi^2_{5} >5.860)=0.32$

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Number: 1, 2 , 3 , 4 , 5 ,6

Frequency: 27, 31, 42, 40, 28, 32

We need to conduct a chi square test in order to check the following hypothesis:

H0: The outcomes are equally likely.

H1: The outcomes are not equally likely.

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The observed values are given:

$O_{1}=27$ $O_{2}=31$

$O_{3}=42$ $O_{4}=40$

$O_{5}=28$ $O_{6}=32$

The expected values are given by:

$E_{1} =\frac{1}{6}*200=33.33$ $E_{2} =\frac{1}{6}*200=33.33$

$E_{3} =\frac{1}{6}*200=33.33$ $E_{4} =\frac{1}{6}*200=33.33$

$E_{5} =\frac{1}{6}*200=33.33$ $E_{6} =\frac{1}{6}*200=33.33$

And now we can calculate the statistic:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

Now we can calculate the degrees of freedom for the statistic given by:

$df=Categories-1=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >5.860)=0.32$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(5.860,5,TRUE)"

4 0

3 years ago

Ella has $115. She wants to purchase a cell phone for $53 and spend the rest on music. Each music album costs $9. Write and solv

lana [24]

53 + 9x < 115
THE LESS THAN SYMBOL SHOULD HAVE A LINE UNDER IT!

3 0

3 years ago

3-74. Mentally calculate the following products. Use the rule for decimal multiplication to write an equation in

dalvyx [7]

B. -.0006

C. .28

Multiply 3 and 2 and you would get 6 then move the decimal 4 places to the left and you would get .0006

Multiply 7 and 4 and you would get 28 then move the decimal places 2 spaces to the left and you would get .28

8 0

3 years ago