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kirill [66]
3 years ago
12

Can you pls help me with these questions? I'm giving brainliest ^^

Mathematics
2 answers:
Eddi Din [679]3 years ago
7 0

Answer:

A. 15, 9, 3, -3, -9

Step-by-step explanation:

That's all I know, sorry

larisa [96]3 years ago
6 0

Answer:

See attached pdf

Step-by-step explanation:

Download pdf
You might be interested in
Figure LMNP will be reflected across the y-axis. Place the point on the graph that represents
zimovet [89]

Answer:

If it's reflected on the y-axis, the answer for N would be (-2,-3).

5 0
3 years ago
It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de
scoray [572]

Answer:

a) P(X

P(z

b) P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)

P(z>-0.583)=1-P(Z

c) P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)

P(z>2.33)=1-P(Z

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) z=1.28

And if we solve for a we got

a=72000 +1.28*12000=87360

So the value of height that separates the bottom 90% of data from the top 10% is 87360.  

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 life of a particular auto transmission of a population, and for this case we know the distribution for X is given by:

X \sim N(72000,12000)  

Where \mu=72000 and \sigma=12000

We are interested on this probability

P(X

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

And we can find this probability using excel or the normal standard table and we got:

P(z

Part b

P(X>65000)

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>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)

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

P(z>-0.583)=1-P(Z

Part c

P(X>100000)

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>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)

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

P(z>2.33)=1-P(Z

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.1   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=1.28

And if we solve for a we got

a=72000 +1.28*12000=87360

So the value of height that separates the bottom 90% of data from the top 10% is 87360.  

5 0
3 years ago
Are these rates equivalent, 96 words typed for 3 minutes; 160 words typed for 5 minutes
monitta
<h2>Greetings</h2>

Answer:

Yes, they are.

Step-by-step explanation:

<h3>1st rate)</h3>

Lets simplify both rates to the rate typer per minute.

(m is minutes)

3m = 96

Divide both sides by 3:

\frac{3m}{3} = \frac{96}{3}

m = 32

So the rate is 32 words per minute.

<h3>2nd rate</h3>

5m = 160

Divide both sides by 5:

\frac{5m}{5} = \frac{160}{5}

m = 32

<h3>So because the rates per minute are the same, they are equivalent.</h3>
<h2>Hope this helps!</h2>
3 0
3 years ago
a worker A finishes a project on his own in 12 days. the worker completes the same work alone in 16 days. the two workers starte
ankoles [38]

Answer:

<em>Worker B needs 9 days to finish the rest of the work</em>

Step-by-step explanation:

<u>Proportions</u>

Worker A finishes a project on his own in 12 days and worker B does the same in 16 days.

Worker A does 1/12 of the project in one day and worker B does 1/16 of the project in one day. When working together, they do

\frac{1}{12}+\frac{1}{16}=\frac{7}{48}

After 3 days they have completed:

3*\frac{7}{48}=\frac{7}{16}

parts of the project, this means it still remains:

1-\frac{7}{16}=\frac{9}{16}

parts of the project and it is done by B alone. Since B does \frac{1}{16} per day, he now takes

\displaystyle \frac{\frac{9}{16}}{\frac{1}{16}}=9

days to complete the project.

Worker B needs 9 days to finish the rest of the work

4 0
3 years ago
A website offers a coupon such that each customer has a
sweet-ann [11.9K]

Answer:

62.29%

Step-by-step explanation:

The probability of Aya being offered a coupon on at least one of the six days she visits the website is 100% minus the probability that she is not offered a coupon on any of the six days, which is described by a binomial probability with zero successes in six trials with a probability of succes p = 0.15.

P = 1 - (\frac{6!}{(6-0)!0!}* p^0*(1-p)^{6-0})\\P = 1 - (1* 1*(1-0.15)^{6})\\P=0.6229=62.29\%

The probability that Aya will be offered a coupon on at least one of the days she visits the website is 62.29%.

8 0
3 years ago
Read 2 more answers
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