I keep getting it wrong help

On Tuesday Clary and Mayla both ran for two and a half hours who ran the farther distance

Ksenya-84 [330]

No one because their is no fractions Or whole numbers or even decimals to show the amount that they ran please check to see if their is any other part to this question

4 0

3 years ago

If there are a total of 25 questions, and I scored 56% on my test, how many answers did I get right, and how many answers did I

Ipatiy [6.2K]

Answer:

14 correct and 11 wrong

Step-by-step explanation:

If you got a 56% , we need to change this to decimal form 56% = .56

This is the percent you got right

.56* 25 = 14

You got 14 questions right

To determine the number of questions you got wrong, take the total number of questions minus the number you got right.

25-14 =11

You got 11 wrong

8 0

2 years ago

\log_{4} x + \log_{4} (x+5) = 6

irina1246 [14]

log(4)x + log(4)(x + 5) = 6
log(4)x + log(4)x + log(20) = 6
 log(8)x + log(20) = 6
 - log(20) - log(20)
 log(8)x = 6 - log(20)
 log(8) log(8)
 x ≈ 0.999

8 0

3 years ago

A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. The standard deviation of the amount in

erica [24]

Answer:

$\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54$

The degrees of freedom are:

$df=n-1=10-1=9$

Now we can calculate the p value using the alternative hypothesis:

$p_v =2*P(\chi^2 >13.54)=0.279$

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

Step-by-step explanation:

Assuming the following data:"12.14 12.05 12.27 11.89 12.06

12.14 12.05 12.38 11.92 12.14"

We can calculate the sample deviation with this formula:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$n=10$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =0.0217$ represent the sample variance

$\sigma^2_0 =0.12^2= 0.0144$ represent the value to verify

Null and alternative hypothesis

We want to determine whether the standard deviation differs from 0.12 ounce, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 0.0144$

Alternative hypothesis: $\sigma^2 \neq 0.0144$

The statistic can be calculated like this;

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

$\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54$

The degrees of freedom are:

$df=n-1=10-1=9$

Now we can calculate the p value using the alternative hypothesis:

$p_v =2*P(\chi^2 >13.54)=0.279$

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

7 0

3 years ago

Please help! I will give brainlest answer

Sidana [21]

Connect the bdc letters together.

8 0

3 years ago

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