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3 years ago

7

Nicole buys envelopes for her home office that come in boxes of 125 envelopes. If she buys 18 boxes, how many envelopes will she

have in all?

1 answer:

yawa3891 [41]3 years ago

6 0

Answer:

➩ 2,250 envelopes.

Step-by-step explanation:

Given that, Nicole buys 125 envelopes per box for her home office.

Therefore, She will have 18*125 envelopes in all

So, 18*125 is 2,250 envelopes

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4 years ago

The Institute of Management Accountants (IMA) conducted a survey of senior finance professionals to gauge members’ thoughts on g

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Answer:

(1) The probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2) The two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3) The two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.

Step-by-step explanation:

Let <em>X</em> = number of senior professionals who thought that global warming is having a significant impact on the environment.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 100 and <em>p</em> = 0.65.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of <em>p</em> if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

Check the conditions as follows:

$np= 100\times 0.65=65>10\\n(1-p)=100\times (1-0.65)=35>10$

Thus, a Normal approximation to binomial can be applied.

So, $\hat p\sim N(p, \frac{p(1-p)}{n})=N(0.65, 0.002275)$ .

(1)

Compute the value of $P(0.64$ as follows:

$P(0.64$

$=P(-0.20$

Thus, the probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2)

Let $p_{1}$ and $p_{2}$ be the two population percentages that will contain the sample percentage with probability 90%.

That is,

$P(p_{1}$

Then,

$P(p_{1}$

$P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$P(-z$

The value of <em>z</em> for P (Z < z) = 0.95 is

<em>z</em> = 1.65.

Compute the value of $p_{1}$ and $p_{2}$ as follows:

$-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.65=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.65\times 0.05)\\p_{1}=0.5675\\p_{1}\approx0.57$ $z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.65=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.65\times 0.05)\\p_{1}=0.7325\\p_{1}\approx0.73$

Thus, the two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3)

Let $p_{1}$ and $p_{2}$ be the two population percentages that will contain the sample percentage with probability 95%.

That is,

$P(p_{1}$

Then,

$P(p_{1}$

$P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$P(-z$

The value of <em>z</em> for P (Z < z) = 0.975 is

<em>z</em> = 1.96.

Compute the value of $p_{1}$ and $p_{2}$ as follows:

$-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.96=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.96\times 0.05)\\p_{1}=0.552\\p_{1}\approx0.55$ $z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.96=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.96\times 0.05)\\p_{1}=0.748\\p_{1}\approx0.75$

Thus, the two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.

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4 years ago

One sample has a mean of M= 6, and a second sample has a mean of M = 12. The two samples are combined into a single set of score

lakkis [162]

Answer:

a) 9

b) 10

c) 8

Step-by-step explanation:

We are given the following in he question:

Mean of sample A = 6

Mean of sample B = 12

Formula:

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$\dfrac{\displaystyle\sum x_a}{n_a} = 6\\\\\dfrac{\displaystyle\sum x_b}{n_b} = 12$

a) original samples have n = 4 scores

$\dfrac{\displaystyle\sum x_a}{4} = 6,\sum x_a = 24\\\\\dfrac{\displaystyle\sum x_b}{4} = 12,\sum x_b = 48\\\\\text{Combined mean} = \displaystyle\frac{\sum x_a + \sum x_b}{4+4}=\frac{24 + 48}{4 + 4} = 9$

b) first sample has n = 3 and the second sample has n = 6

$\dfrac{\displaystyle\sum x_a}{3} = 6,\sum x_a = 18\\\\\dfrac{\displaystyle\sum x_b}{6} = 12,\sum x_b = 72\\\\\text{Combined mean} = \displaystyle\frac{\sum x_a + \sum x_b}{3+6}=\frac{18+72}{3+6} = 10$

c) first sample has n = 6 and the second sample has n = 3

$\dfrac{\displaystyle\sum x_a}{6} = 6,\sum x_a = 36\\\\\dfrac{\displaystyle\sum x_b}{3} = 12,\sum x_b = 36\\\\\text{Combined mean} = \displaystyle\frac{\sum x_a + \sum x_b}{6+3}=\frac{36+36}{6+3} = 8$

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