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AveGali [126]
2 years ago
12

Pls help me!!!!solve this question

Mathematics
1 answer:
nasty-shy [4]2 years ago
3 0
(11k^4 - 4m) + (-2k^4 - 15m)
Add -2k^4 to 11k^4. This would be the same as subtracting 2k^4 from 11k^4
9k^4 - 4m - 15m
Add -15m to -4m. This would be the same as subtracting 15m from -4m
Final Answer: 9k^4 - 19m

Word Form: Nine times the variable k to the forth power minus nineteen times the variable m.
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Express M=cos3A + cos A and N = sin3A + sin A as a product form​
elena-14-01-66 [18.8K]

Answer:

2cosAcos2A, 4sinAcos^2A

Step-by-step explanation:

cos3A+cosA

2cos((3A+A)/2)cos((3A-A)/2)

2cos(4A/2)cos(2A/2)

2cosAcos2A

sin3A+sinA

2sin((3A+A)/2)cos((3A-A)/2)

2sin(4A/2)cos(2A/2)

2sin2AcosA

4sinAcos^2A

3 0
2 years ago
The ratio of blue skittles to red skittles in Tayler's bag is 3 to 4. If there are 20 red
adoni [48]

Answer:

Tayler has 35 SHKITTLES IN ALL

Step-by-step explanation:

Since Tayler has 20 red Skittles, and the ratio of blue skittles to red skittles is 3 to 4, that means that Tayler has 3/4 as many blue skittles as she does red skittles. 3/4 of 20 is 15

20 + 15 = 35

Thus, Tayler has 35 skittles

SHKITTLES lol

6 0
2 years ago
49.5 is what percent of 33?
stira [4]
49.5/33 x 100= 150%. This is also rather simple to figure out just by looking at the question itself, you can clearly see that 49.5 is 16.5 greater than 33 and that 16.5 is half of 33. Which you can easily figure out from there.
6 0
3 years ago
The Insurance Institute reports that the mean amount of life insurance per household in the US is $110,000. This follows a norma
nata0808 [166]

Answer:

a) \sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85

b) Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

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

c) P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)

And we can use the complement rule and we got:

P(Z>0.354) = 1-P(Z

d) P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)

And we can use the complement rule and we got:

P(Z>-1.768) = 1-P(Z

e) P(100000< \bar X

And we can use the complement rule and we got:

P(-1.768

Step-by-step explanation:

a. If we select a random sample of 50 households, what is the standard error of the mean?

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 the amount of life insurance of a population, and for this case we know the distribution for X is given by:

X \sim N(110000,40000)  

Where \mu=110000 and \sigma=40000

If we select a sample size of n =35 the standard error is given by:

\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85

b. What is the expected shape of the distribution of the sample mean?

Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

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

c. What is the likelihood of selecting a sample with a mean of at least $112,000?

For this case we want this probability:

P(X > 112000)

And we can use the z score given by:

z= \frac{\bar X  -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)

And we can use the complement rule and we got:

P(Z>0.354) = 1-P(Z

d. What is the likelihood of selecting a sample with a mean of more than $100,000?

For this case we want this probability:

P(X > 100000)

And we can use the z score given by:

z= \frac{\bar X  -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)

And we can use the complement rule and we got:

P(Z>-1.768) = 1-P(Z

e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

For this case we want this probability:

P(100000

And we can use the z score given by:

z= \frac{\bar X  -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

P(100000< \bar X

And we can use the complement rule and we got:

P(-1.768

8 0
3 years ago
Ana drinks chocolate milk out of glasses that each hold 1/8 of a liter. She has 3/4 a liter of chocolate milk in her refrigerato
Zigmanuir [339]
There are 6 eights in 3/4, so six

7 0
3 years ago
Read 2 more answers
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