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2 years ago

9

Lily has 15 shells in her shell collection. Krystal has twice as many shells in her collection as lily does. Mara has 11 fewer s

hells than Krystal. How many shells does Mara have?

2 answers:

tester [92]2 years ago

6 0

Mara has 19 shells. I hoped this helped! Mark as branliest if possible!

Hitman42 [59]2 years ago

6 0

I think that mara have 4 shells

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Find the length of the arc intercepted by a central angle calculator in pi

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Give me a screenshot

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3 years ago

What is the value of x in <br> 150-35x=45

Annette [7]

Your answer is x = 3 because if you take 150 and subtract it by both sides:
150 - 35x = 45
-150 -150

you have to cross out both 150 and -150, take 45 and subtract 150 which will give you -105.
-35x = -105

take -35 and divide by both sides. Cross out the two -35's. Take -105/-35 which your answer is x = 3. Hopes this helps. :)

~Shadow

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3 years ago

The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 21 people reveals the mean yearly consum

PIT_PIT [208]

Correct question is;

The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 21 people reveals the mean yearly consumption to be 74 gallons with a standard deviation of 16 gallons.

a. What is the value of the population mean? What is the best estimate of this value?

b. Explain why we need to use the t distribution. What assumption do you need to make?

c. For a 90 percent confidence interval, what is the value of t?

d. Develop the 90 percent confidence interval for the population mean.

e. Would it be reasonable to conclude that the population mean is 68 gallons?

Answer:

A) Best estimate = 74 gallons

B) because the population standard deviation is unknown. The assumption we will make is that the population follows the normal distribution.

C) t = 1.725

D) 90% confidence interval for the population mean is (67.9772, 80.0228) gallons

E) Yes

Step-by-step explanation:

We are given;

Sample mean; x' = 74

Sample population; n = 21

Yearly Standard deviation; s = 16

A) We are not given the population mean.

So the closest estimate to the population mean would be the sample mean which is 74.

B) We are not given the population standard deviation and as such we can't use normal distribution. So what is used when population standard deviation is not known is called t - distribution table. The assumption we will make is that the population follows the normal distribution.

C) At confidence interval of 90% and DF = n - 1 = 21 - 1 = 20

From t-tables, the t = 1.725

D) Formula for the confidence interval is;

x' ± t(s/√n) = 74 ± 1.725(16/√21) = 74 ± 6.0228 = 67.9772 or 80.0228

Thus 90% confidence interval for the population mean is (67.9772, 80.0228) gallons

E) 68 gallons lies within the range of the confidence interval, thus we can say that "Yes, it is reasonable"

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3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

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3 years ago

What is -3 (6x -4) + 2 (2x -4) = - 50

padilas [110]

Answer:

$\frac{27}{7}$

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2 years ago

Read 2 more answers