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3 years ago

Brennon has a summer job working for the

1 answer:

6 0

Brennon should make about $212.05. since 153/18 is 8.5, all we do is multiply 8.5 by 25. that gives us 212.5, convert it to money and you get $212.05.

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A simple random sample of size nequals=8181 is obtained from a population with mu equals 77μ=77 and sigma equals 27σ=27. (a) De

ivanzaharov [21]

Answer:

a) $P(\bar X>81.5)=1-0.933=0.067$

b) $P(\bar X$

c) $P(73.4$

Step-by-step explanation:

1) 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".

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

$X \sim N(\mu=77,\sigma=27)$

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

On this case $\bar X \sim N(77,\frac{27}{\sqrt{81}})$

Part a

We want this probability:

$P(\bar X>81.5)=1-P(\bar X$

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

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

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

$P(\bar X >81.5)=1-P(Z$

$P(\bar X>81.5)=1-0.933=0.067$

Part b

We want this probability:

$P(\bar X\leq 69.5)$

If we apply the formula for the z score to our probability we got this:

$P(\bar X \leq 69.5)=P(Z\leq \frac{69.5-77}{\frac{27}{\sqrt{81}}})=P(Z$

$P(\bar X\leq 69.5)=0.0062$

Part c

We are interested on this probability

$P(73.4$

If we apply the Z score formula to our probability we got this:

$P(73.4$

$=P(\frac{73.4-77}{\frac{27}{\sqrt{81}}}$

And we can find this probability on this way:

$P(-1.2$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(-1.2$

3 0

4 years ago

(-2,11)slope -4 write the linear equation

kodGreya [7K]

Answer:

y = − 4 x + 3

Step-by-step explanation:

8 0

3 years ago

Select the correct difference.

scZoUnD [109]

Answer: First option is correct.

Step-by-step explanation:

Since we have given that

$5d^2+4d-3-(2d^2-3d+4)$

We need to find the correct difference.

Collect the like terms together.

$5d^2+4d-3-(2d^2-3d+4)\\\\=5d^2+4d-3-2d^2+3d-4\\\\=5d^2-2d^2+4d+3d-3-4\\\\=3d^2+7d-7$

Hence, First option is correct.

5 0

3 years ago

Read 2 more answers

Do the table and the equation represent the same function ? Y=390+11(x)

Furkat [3]

<u>Answer:</u>

<u>Step-by-step explanation:</u>

We are given the following equation of a function and a table for the corresponding values of this function:

$y=390+11(x)$

We are to determine if the equation and the table represent the same function.

To check that, we will substitute the value of x in the equation to see if it gives the same values of y as in the table.

$y=390+11(-30)$ ---> (-30, 60)

$y=390+11(-28)$ ---> (-28, 82)

$y=390+11(-26)$ ---> )-26, 104)

Since these paired values differ from the ones given in the table. Therefore, table and equation do not represent the same function.

8 0

4 years ago

On average 25% of dogs who go to ABC veterinary need a rabies booster. If 120 dogs visit ABC vet Marion veterinarian how many of

Daniel [21]

Answer:

The no. of digs needs a rabies booster = 30

Step-by-step explanation:

Given data

Total no. of dogs visit ABC vet Marion veterinarian = 120

$No. \ of \ dogs \ needs \ a \ rabies \ booster = \frac{25}{100} Total \ no \ of \ dogs \ who \ go \ to \ ABC \ veterinary.$ $No. \ of \ dogs \ needs \ a \ rabies \ booster = \frac{25}{100} (120)$

$No. \ of \ dogs \ needs \ a \ rabies \ booster = 30$

Therefore the no. of digs needs a rabies booster = 30

7 0

4 years ago