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3 years ago

Hii please help i’ll give brainliest

Mathematics

2 answers:

ivanzaharov [21]3 years ago

6 0

C

1) 2.25
2) 0.35
3) -2.25
4) -0.25

Zepler [3.9K]3 years ago

3 0

Answer:

C- the third one

Step-by-step explanation:

It is the only one that shows the dot on -2

if this helps, brainliest, please?????

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Select the best definitions of population and sample. A sample is the group from whom information is being collected. A populati

Elena L [17]

Answer:

A population is defined as the data set which consists of all members of a specified group and the sample contains a part or a subset of a population.

For example : all the students of a school are population. But out of all, how many students play sports is a sample.

Hence, the answer is - A population is the complete group under study. A sample is the sub-collection of members of the population from which data are actually collected.

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4 years ago

Please help !! Find the value of x

Naily [24]

Answer:

Step-by-step explanation:

6x-3+75=180

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6x=108

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3 years ago

HURRY!!!!!!!!!!!

andrey2020 [161]

Answer:

2.5

Step-by-step explanation:

Conversion a mixed number 2 1/

to a improper fraction: 2 1/2 = 2 1/

= 2 · 2 + 1/

= 4 + 1/

= 5/

To find new numerator:

a) Multiply the whole number 2 by the denominator 2. Whole number 2 equally 2 * 2/

= 4/

b) Add the answer from previous step 4 to the numerator 1. New numerator is 4 + 1 = 5

c) Write a previous answer (new numerator 5) over the denominator 2.

Two and one half is five halfs

4 0

3 years ago

Read 2 more answers

Solve for C:<br> Spam: deiodhwidweiudjegydjwe

Maksim231197 [3]

Answer:

$C=\frac{5F-160}{9}$

Step-by-step explanation:

Let's solve for C.

$F=\frac{9C}{5}$

Step 1: Flip the equation.

$\frac{9}{5} C+32=F$

Step 2: Add $-32$ to both sides.

$\frac{9}{5} C+32+-32=F+-32$

$\frac{9}{5}C=F-32$

Step 3: Divide both sides by $\frac{9}{5}$ .

$\frac{\frac{9}{5}C}{\frac {9} {5}}=\frac{F-32}{\frac{9}{5} }$

$C=\frac{5F-160}{9}$

Answer:

$C=\frac{F-160}{9}$

8 0

3 years ago

Experian would like to test the hypothesis that the average credit score for an adult in Virginia is different from the average

aliya0001 [1]

Answer:

a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

Step-by-step explanation:

Let $\mu_{1}-\mu_{2}$ be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes $n_{1} = 40$ and $n_{2} = 35$ , the unbiased point estimate for $\mu_{1}-\mu_{2}$ is $\bar{x}_{1} - \bar{x}_{2}$ , i.e., 699-682 = 17.

The standard error is given by $\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$ , i.e.,

$\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}}$ = 9.8198.

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative). The rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is $Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}$ and the observed value is $z_{0} = \frac{17}{9.8198} = 1.7312$ . Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.

b. The endpoints for a 95% confidence interval for $\mu_{1}-\mu_{2}$ is given by $17\pm (z_{0.05/2})9.8198$ , i.e., $17\pm (z_{0.025})9.8198$ where $z_{0.025}$ is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

3 0

4 years ago