Write an equivalent expression. 2 + 4(c + 7)

tara has two bags of marbles. the first bag contains 6 red marbles , 5 blue marbles and 4 green marbles tara will recomendly sel

vodka [1.7K]

well, in order to solve this we can put each marble into a fraction.

so for blue it would be 5 over 15 (5/15)

because there are 15 marbles in the bag but only 5 of them are blue!

but if you need a % then just divide the top by bottom

then you get a 33% chance of picking blue! hope this helps!

4 0

4 years ago

A researcher collected data from a sample of correctional officers and found that on average they made 47,173 dollars a year wit

ollegr [7]

Answer:

a) $z = \frac{40000- 47173}{6364} = -1.127$

b) $z = \frac{50000- 47173}{6364} = 0.444$

c) $P(X>40000)=P(\frac{X-\mu}{\sigma}>\frac{40000-\mu}{\sigma})=P(Z>\frac{40000-47173}{6364})=P(z>-1.127)$

And we can find this probability using the complement rule and the normal standard distribution or excel and we got:

$P(z>-1.127)=1-P(z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the amount of money of a population of interet, and we know the following info:

Where $\bar X=47173$ and $s=6364$

We can assume that the sample size is large and the estimators can be used as a good description for the parameters $\mu \sigma$

The z score for 40000 would be:

$z = \frac{40000- 47173}{6364} = -1.127$

Part b

The z score for 50000 would be:

$z = \frac{50000- 47173}{6364} = 0.444$

Part c

We are interested on this probability

$P(X>40000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>40000)=P(\frac{X-\mu}{\sigma}>\frac{40000-\mu}{\sigma})=P(Z>\frac{40000-47173}{6364})=P(z>-1.127)$

And we can find this probability using the complement rule and the normal standard distribution or excel and we got:

$P(z>-1.127)=1-P(z$

8 0

3 years ago

NEED HELP ASAP!!!!!

julsineya [31]

B, C, and D
Hope this helped:)

7 0

3 years ago

Read 2 more answers

25 students in a high school health class are asked to write down what foods they ate last week. Of the 25

kondor19780726 [428]

Answer:

14/25, 6/25, 1/5, 3/5

Step-by-step explanation:

5 0

3 years ago

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne

Pavlova-9 [17]

Answer:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

$p_v =P(t_{(15)}>1.507)=0.076$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{CTS}=2.35$ represent the mean for the sample CTS

$\bar X_{N}=1.83$ represent the mean for the sample Normal

$s_{CTS}=0.88$ represent the sample standard deviation for the sample of CTS

$s_{N}=0.54$ represent the sample standard deviation for the sample of Normal

$n_{CTS}=10$ sample size selected for the CTS

$n_{N}=7$ sample size selected for the Normal

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis: $\mu_{CTS} \leq \mu_{N}$

Alternative hypothesis: $\mu_{CTS} > \mu_{N}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{CTS}+n_{N}-2=10+7-2=15$

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(15)}>1.507)=0.076$

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

6 0

4 years ago