How to right two hundred seven thousand forty eight in standard form

4. What is the slope of the line that passes through the given points? (3, 2) and (5, 12)

valina [46]

Answer:5

Step-by-step explanation:

y2-y1/x2-x1

12-2/5-3

10/2

=5

8 0

3 years ago

AKS 18/19 A travel agent is arranging a private cruise. The agent can arrange for, at most, 40 people to go on the trip. The tri

zlopas [31]

Answer:

M + W <= 40

M >= 12

M >= 15

Total profit = $26 * M + $14 * W

Step-by-step explanation:

Let the number of men be denoted by M

number of women be denoted by W

Maximum no of people that can go one trip = 40

Therefore representing it mathematically we have

M + W <= 40 -------------- CONSTRAINT 1

Minimum number of men needed on trip = 12

Therefore representing it mathematically we have

M >= 12 CONSTRAINT 2

Minimum number of women needed on trip = 15

Therefore representing it mathematically we have

M >= 15 CONSTRAINT 3

profit on 1 man = $26

profit on M men = $26 * M

profit on 1 woman = $14

profit on M men = $14 * W

Total profit combining for men and women =

$26 * M + $14 * W -------------CONSTRAINT 4

___________________________________________

writing together all the constraint

M + W <= 40

M >= 12

M >= 15

Total profit = $26 * M + $14 * W

4 0

3 years ago

X* 6 = 18; Solve for x. *

Digiron [165]

Answer:

x = 3

Step-by-step explanation:

x*6 = 6x, then we divide both sides by 6 to get x alone. 6x/6 = 18/6.

x = 3, 18 divided by 6 is 3.

4 0

3 years ago

Tyler earned a grade of 88% on a math test that had 25 problems. How many problems on this test did she answer incorrectly?

kolbaska11 [484]

Answer:

22

Step-by-step explanation:

3 0

3 years ago

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

DochEvi [55]

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

The random variable X follows a Binomial distribution with parameters n = 100 and p = 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 p 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 z for P (Z < z) = 0.95 is

z = 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 z for P (Z < z) = 0.975 is

z = 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.

7 0

4 years ago