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2 years ago

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A large bag of marbles contains red (R) and blue (B) marbles. Marbles are pulled one at a time, recorded, and placed back in the

bag. Estimate the percentage of blue marbles in the bag.
R R R B R R R R R R B R R R B

A. 60%

B. 80%

C. 25%

D. 20%

1 answer:

ehidna [41]2 years ago

6 0

The answer will be A.

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Mateus’s bank issued an advertisement saying that 90\%90%90, percent of its customers are satisfied with the bank’s services. Si

densk [106]

Answer:

The probability of getting a sample with 80% satisfied customers or less is 0.0125.

Step-by-step explanation:

We are given that the results of 1000 simulations, each simulating a sample of 80 customers, assuming there are 90 percent satisfied customers.

Let $\hat p$ = <u><em>sample proportion of satisfied customers</em></u>

The z-score probability distribution for the sample proportion is given by;

Z = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, p = population proportion of satisfied customers = 90%

n = sample of customers = 80

Now, the probability of getting a sample with 80% satisfied customers or less is given by = P( $\hat p \leq$ 80%)

P( $\hat p \leq$ 80%) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ $\leq$ $\frac{0.80-0.90}{\sqrt{\frac{0.80(1-0.80)}{80} } }$ ) = P(Z $\leq$ -2.24) = 1 - P(Z < 2.24)

= 1 - 0.9875 = <u>0.0125</u>

The above probability is calculated by looking at the value of x = 2.24 in the z table which has an area of 0.9875.

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2 years ago

ASAP PLEASE HELP!!

jenyasd209 [6]

Answer:

save up and pay cash

Step-by-step explanation:

paying cash is always the best idea and not trying to get so much debit. If you did one other otion and have the debit some emergency can come up and you will not be able to pay for that, and aying cash you save more money

5 0

2 years ago

Read 2 more answers

se the function to show that fx(0, 0) and fy(0, 0) both exist, but that f is not differentiable at (0, 0). f(x, y) = 9x2y x4 + y

alexandr1967 [171]

Answer:

It is proved that $f_x, f_y$ exixts at (0,0) but not differentiable there.

Step-by-step explanation:

Given function is,

$f(x,y)=\frac{9x^2y}{x^4+y^2}; (x,y)\neq (0,0)$

To show exixtance of $f_x(0,0), f_y(0,0)$ we take,

$f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0$ exists.

And,

$f_y(0,0)=\lim_{k\to 0}\frac{f(h,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{9h^2k}{k(h^4+k^2)}=\lim_{k\to 0}\frac{9h^2}{h^4+k^2}=\frac{9}{h^2}$ exists.

To show f(x,y) is not differentiable at the origin cheaking continuity at origin be such that,

$\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}$ where m is a variable.

which depends on various values of m, therefore limit does not exists. So f(x,y) is not continuous at (0,0). Hence it is not differentiable at (0,0).

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2 years ago

Please help me with this question #8

MA_775_DIABLO [31]

Think of sqrt(2) as a variable.

3x - x = 2x

If x = sqrt(2), then answer is 2sqrt(2)

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2 years ago

Julie knows that the adult population gets, on average, eight hours of sleep each night. A hypothesis test can help her see if c

Mamont248 [21]

Answer:

-3.64

Step-by-step explanation:

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2 years ago