Which expressions are equivalent?

fathi has $1.10 in his printing account. each sheet of paper he uses reduces his printing account balance by $0.25. Fathi wants

Oksana_A [137]

Answer:

IF THE question is After Fathi prints, what will be the balance in his printing account?

then it is: $1.9

There are 47 pages.

Printing on both sides would divide the number of pages into half.

47/2 = 23.5

2 pages on each side would mean 4 pages on one sheet. Therefore, the number of pages will be further divided by 2.

23.5/2 = 11.75

There cannot be 11.75 pages so we will round it up to 12 pages.

Each page costs $0.25 so 12 pages will cost:

12 x 0.25 = $3

Faithi has $1.1 so new account balance will be:

1.1 - 3 = $-1.9

Therefore, Fathi's balance in his printing account would be negative $1.9.

6 0

3 years ago

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Suppose that a basketball player can score on a particular shot with probability .3. Use the central limit theorem to find the a

Rom4ik [11]

Answer:

(a) The probability that the number of successes is at most 5 is 0.1379.

(b) The probability that the number of successes is at most 5 is 0.1379.

(c) The probability that the number of successes is at most 5 is 0.1379.

(d) The probability that the number of successes is at most 11 is 0.9357.

→ All the exact probabilities are more than the approximated probability.

Step-by-step explanation:

Let S = a basketball player scores a shot.

The probability that a basketball player scores a shot is, P (S) = p = 0.30.

The number of sample selected is, n = 25.

The random variable $S\sim Bin(25,0.30)$

According to the central limit theorem if the sample taken from an unknown population is large then the sampling distribution of the sample proportion ( $\hat p$ ) follows a normal distribution.

The mean of the the sampling distribution of the sample proportion is: $E(\hat p)=p=0.30$

The standard deviation of the the sampling distribution of the sample proportion is:

$SD(\hat p)=\sqrt{\frac{ p(1- p)}{n} }=\sqrt{\frac{ 0.30(1-0.30)}{25} }=0.092$

(a)

Compute the probability that the number of successes is at most 5 as follows:

The probability of 5 successes is: $p=\frac{5}{25} =0.20$

$P(\hat p\leq 0.20)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.20-0.30}{0.092} )\\=P(Z\leq -1.087)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 5 is 0.1379.

The exact probability that the number of successes is at most 5 is:

$P(S\leq 5)={25\choose 5}(0.30)^{5}91-0.30)^{25-5}=0.1935$

The exact probability is more than the approximated probability.

(b)

Compute the probability that the number of successes is at most 7 as follows:

The probability of 5 successes is: $p=\frac{7}{25} =0.28$

$P(\hat p\leq 0.28)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.28-0.30}{0.092} )\\=P(Z\leq -0.2174)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 7 is 0.4129.

The exact probability that the number of successes is at most 7 is:

$P(S\leq 57)={25\choose 7}(0.30)^{7}91-0.30)^{25-7}=0.5118$

The exact probability is more than the approximated probability.

(c)

Compute the probability that the number of successes is at most 9 as follows:

The probability of 5 successes is: $p=\frac{9}{25} =0.36$

$P(\hat p\leq 0.36)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.36-0.30}{0.092} )\\=P(Z\leq 0.6522)\\=0.7422$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 9 is 0.7422.

The exact probability that the number of successes is at most 9 is:

$P(S\leq 9)={25\choose 9}(0.30)^{9}91-0.30)^{25-9}=0.8106$

The exact probability is more than the approximated probability.

(d)

Compute the probability that the number of successes is at most 11 as follows:

The probability of 5 successes is: $p=\frac{11}{25} =0.44$

$P(\hat p\leq 0.44)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.44-0.30}{0.092} )\\=P(Z\leq 1.522)\\=0.9357$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 11 is 0.9357.

The exact probability that the number of successes is at most 11 is:

$P(S\leq 11)={25\choose 11}(0.30)^{11}91-0.30)^{25-11}=0.9558$

The exact probability is more than the approximated probability.

6 0

3 years ago

Midpoint. (3,3), (-1,1)

Scilla [17]

Answer:

1,2

Step-by-step explanation:

3+-1=2

2/2=1

3+1=4

4/2=2

7 0

2 years ago

What is the decay factor of the exponential function represented by the table?

bulgar [2K]

Given:

The table of values of an exponential function.

To find:

The decay factor of the exponential function.

Solution:

The general form of an exponential function is:

$y=ab^x$ ...(i)

Where, a is the initial value and $0$ is the decay factor and $b>1$ is the growth factor.

The exponential function passes through the point (0,6). Substituting $x=0,y=6$ in (i), we get

$6=ab^0$

$6=a(1)$

$6=a$

The exponential function passes through the point (1,2). Substituting $x=1,y=2,a=6$ in (i), we get

$2=6(b)^1$

$2=6b$

$\dfrac{2}{6}=b$

$\dfrac{1}{3}=b$

Here, $b=\dfrac{1}{3}$ lies between 0 and 1. Therefore, the decay factor of the given exponential function is $\dfrac{1}{3}$ .

Hence, the correct option is A.

8 0

2 years ago

What are the correct steps for solving the following equation: 5x - 4= 21

AysviL [449]

1. Add 4 to both sides of the equation
2. Simplify
3. Divide both sides of the equation by 5
4. Simplify
5. x= 5

5 0

2 years ago

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