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2 years ago

List the factors in the expression 4r + 5s +8. Plz help me.

Mathematics

1 answer:

beks73 [17]2 years ago

5 0

Answer: r = -2.4

Step-by-step explanation:

Simplifying 6r + 4 = r + -8 Reorder the terms: 4 + 6r = r + -8 Reorder the terms: 4 + 6r = -8 + r Solving 4 + 6r = -8 + r Solving for variable 'r'. -8 + -4 = -12

5r = -12

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Evaluate the summation of negative 5 n minus 1, from n equals 3 to 12..

NARA [144]

The summation of negative 5 n minus 1, from n equals 3 to 12 can be expressed as

n = 3/12

substitute the value of n to the equatio

-5n – 1

-5( 3/12 ) -1

-9/4

5 0

3 years ago

Pollsters are concerned about declining levels of cooperation among persons contacted in surveys. A pollster contacts 99 people

sertanlavr [38]

Answer:

The probability of getting someone in the 18-21 age bracket or someone who agreed to respond $= 0.365$

Step-by-step explanation:

The counts are as follows

Age Group No of people contacted No of refusals Total

18-21 88 8 96

22-29 240 42 282

Total 328 50 378

The probability of getting some one in the age bracket of 18-21 or some one who refused to respond is

$= \frac{96}{378} + \frac{50}{378} - \frac{8}{378} \\= \frac{96+50-8}{378} \\= \frac{138}{378} \\= 0.365$

8 0

3 years ago

It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken to estimate

Hitman42 [59]

Answer:

The minimum sample size is n = 75 so that the desired margin of error is 5 or less.

Step-by-step explanation:

We are given the following in the question:

Population variance = 484

Standard deviation =

$\sigma^2 = 484\\\sigma =\sqrt{484} = 22$

Confidence level = 0.95

Significance level = 0.05

Margin of error = 5

Formula:

Margin of error =

$E = z\times \dfrac{\sigma}{\sqrt{n}}\\\\n = (\dfrac{z\times \sigma}{E})^2$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get

$E\leq 5\\\\1.96\times \dfrac{22}{\sqrt{n}} \leq 5\\\\\sqrt{n} \geq 1.96\times \dfrac{22}{5}\\\\n \geq (1.96\times \dfrac{22}{5})^2\\\\n \geq 74.373$

Thus, the minimum sample size is n = 75 so that the desired margin of error is 5 or less.

3 0

2 years ago

In a school of 330 students, 85 of them are in a drama club, 200 of them are in a sports team and 60 of them do both

Neko [114]

Answer:

105/330 = 21/66 = 7/22

7/22 is the answer I think

Step-by-step explanation:

n(U) = 330

n(A) = 85

n(B) = 200

n(A n B) = 60

Now,

n(A U B) = n(A) + n(B) - n(A n B)

or, n(A U B) = 85 + 200 - 60

or, n(A U B) = 285 - 60 = 225

Now,

n(A U B) compliment = n(U) - n(A U B)

n(A U B) compliment = 330 - 225

so, n(A U B) compliment = 105

Now,

probability = 105/330

8 0

2 years ago

the times that gym members spend on a treadmill are normally distributed with a mean of 20 min and a standard deviation of 8 min

Papessa [141]

<span>Given:
xbar = 20 min
standard deviation = 8 min
total sample = 240

Required:
how many gym members are expected to spend 4 min and 36 min on a treadmill today

Solution:
sd</span>² = (∑(x - xbar)²) / N - 1

there are 42 gym members at 4 minutes and 200 gym members in 36 minutes.

6 0

3 years ago