Ask question

Ask question

All categories

3 years ago

6

Nancy is allowed to make 200 copies a month on the departmental copy machine. A test for one class requires 26 copies of two pag

es , and a test for a second class requires 21 copies of two pages , how many copies will she have left for the rest of the month after these copies are made?

1 answer:

alekssr [168]3 years ago

6 0

She would have 106 copies left.I think this because since she has to make 47 copies and since they are double sided you would have to multiply be 2.Which would give you 94 and then you subtract 94 from 200.It would give you 106 in the end.

You might be interested in

Need help with this

Alecsey [184]

$\frac{x {}^{3} - 3x {}^{2} - 10x {}^{2} + 30x + 4x - 12 }{x - 3 } \\ \frac{x {}^{2} \times (x - 3) - 10x \times (x - 3) + 4 \times (x - 3) }{x - 3} \\ \frac{(x - 3)(x {}^{2} - 10x + 4 )}{x - 3} \\ x {}^{2} - 10x + 4$

8 0

3 years ago

Spring stretches by 21.0 cm when a 135 n object is attached. what is the weight of a fish that would stretch the spring by 62.1

Marta_Voda [28]

Caution: you need to use the same units of measurement throughout. If the spring stretches by 21 cm when a 135 newton object is attached, then you must ask for the mass (in newtons) of a fish that would stretch the spring by 62.1 cm.

We will need to assume that the spring is not stretched at all if and when no object is attached to the spring.

Write the ratio

21.0 cm 135 newtons
------------- = --------------------
62.1 cm x

Solve this for x. This x value represents the mass of a fish that would stretch the spring by 62.1 cm. You can cancel "cm" in the equation above:

21.0 135 newtons
------ = --------------------
62.1 x

Then 21.0x = (62.1)(135 newtons). Divide both sides of this equation by 21.0 to solve it for x.

7 0

3 years ago

Read 2 more answers

Match each function with its function. Use function composition to determine your answer

Ket [755]

Answer:

C

Step-by-step explanation:

idk

8 0

3 years ago

According to a Yale program on climate change communication survey, 71% of Americans think global warming is happening.† (a) For

SpyIntel [72]

Answer:

a) 0.2741 = 27.41% probability that at least 13 believe global warming is occurring

b) 0.7611 = 76.11% probability that at least 110 believe global warming is occurring

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$p = 0.71$

(a) For a sample of 16 Americans, what is the probability that at least 13 believe global warming is occurring?

Here $n = 16$ , we want $P(X \geq 13)$ . So

$P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 13) = C_{16,13}.(0.71)^{13}.(0.29)^{3} = 0.1591$

$P(X = 14) = C_{16,14}.(0.71)^{14}.(0.29)^{2} = 0.0835$

$P(X = 15) = C_{16,15}.(0.71)^{15}.(0.29)^{1} = 0.0273$

$P(X = 16) = C_{16,16}.(0.71)^{16}.(0.29)^{0} = 0.0042$

$P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741$

0.2741 = 27.41% probability that at least 13 believe global warming is occurring

(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?

Now $n = 160$ . So

$\mu = E(X) = np = 160*0.71 = 113.6$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{160*0.71*0.29} = 5.74$

Using continuity correction, this is $P(X \geq 110 - 0.5) = P(X \geq 109.5)$ , which is 1 subtracted by the pvalue of Z when X = 109.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{109.5 - 113.6}{5.74}$

$Z = -0.71$

$Z = -0.71$ has a pvalue of 0.2389

1 - 0.2389 = 0.7611

0.7611 = 76.11% probability that at least 110 believe global warming is occurring

3 0

3 years ago

2. Solve this equation for x:<br> 0.17k - 0.43 = 0.25k + 0.05

stepan [7]

Answer:

k = 6

Step-by-step explanation:

im guessing you meant solve for k

0.17k - 0.43 = 0.25k + 0.05

-0.17k -0.17k

-0.43 = 0.08k + 0.05

-0.05 -0.05

-0.48 = 0.08k

divide both sides of the equal sign by 0.08

6 = k

4 0

3 years ago