Find the quotient and remainder using long division.x + x - 13
x - 2
The quotient is
The remainder is

Answers

Answer 1

Answer:

Answer:

$\text{The quotient is}~ x^2 +2x +5 \n\n\text{The remainder is}~ -3$

Step-by-step explanation:

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A professor graded the final exams and found that the mean score was 70 points. Which of the following can you conclude?A- All of the above.B- The median score was 70 points.C- 50% of the students scored below 70 points.D- This would be a normal distribution.
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On an exam, the mean score is 78 points, with a standard deviation of 6 points. Assuming normal distribution of the scores, approximately what percentage of students received more than 85?A.12% B.17% C.8% D.9% E.None of the above

Answers

Answer: A. 12%

Step-by-step explanation:-

Given : In an exam , Mean score : $\mu=78\text{ points}$

Standard deviation : $6\text{ points}$

Let X be a random variable that represents the scores of students.

We assume that the points are normally distributed.

Z-score : $z=(x-\mu)/(\sigma)$

For x = 85, we have

$z=(85-78)/(6)\approx1.17$

Then using standard normal distribution table, the probability that the students received more than 85 is given by :-

$P(x>85)=P(z>1.17)=1-P(z<1.17)\n\n=1-0.8789995=0.1210005\approx0.12=12\%$

Hence, the percentage of students received more than 85 =12%

You are trying to determine if you should accept a shipment of eggs for a local grocery store. About 4% of all cartons which are shipped have had an egg crack while traveling. You are instructed to accept the shipment if no more than 10 cartons out of the 300 you inspect have cracked eggs. What is the probability that you accept the shipment? (In other words, what is the probability that, at the most, you had 16 cartons with cracked eggs?)a. 1012.
b. 3669.
c. 5319.
d. 4681.
e. 5832.
f. 6331.

Answers

Answer:

0.8809

Step-by-step explanation:

Given that:

The population proportion p = 4% = 4/100 = 0.04

Sample mean x = 16

The sample size n = 300

The sample proportion $\hat p =(x)/(n)$

= 16/300

= 0.0533

∴

$P(\hat p \leq 0.0533) = P\bigg ( \frac{\hat p - p}{\sqrt{(P(1-P))/(n)}}\leq\frac{0.0533 - 0.04}{\sqrt{(0.04(1-0.04))/(300)}}\bigg )$

$P(\hat p \leq 0.0533) = P\bigg ( Z\leq\frac{0.0133}{\sqrt{(0.04(0.96))/(300)}}\bigg )$

$P(\hat p \leq 0.0533) = P\bigg ( Z\leq\frac{0.0133}{\sqrt{(0.0384)/(300)}}\bigg )$

$P(\hat p \leq 0.0533) = P\bigg ( Z\leq\frac{0.0133}{\sqrt{1.28 * 10^(-4)}}\bigg )$

$P(\hat p \leq 0.0533) = P\bigg ( Z\leq(0.0133)/(0.0113)}\bigg )$

$P(\hat p \leq 0.0533) = P\bigg ( Z\leq1.18}\bigg )$

From the z tables;

= 0.8809

Let X be the random variation that follows a normal distribution;

Then;

population mean $\mu$ = n × p

population mean $\mu$ = 300 × 0.04

population mean $\mu$ = 12

The standard deviation $\sigma = √(np(1-p))$

The standard deviation $\sigma = √(300 * 0.04(1-0.04))$

The standard deviation $\sigma = √(11.52)$

The standard deviation $\sigma = 3.39$

The z -score can be computed as:

$z = (x - \mu)/(\sigma)$

$z = (16 -12)/(3.39)$

$z = (4)/(3.39)$

z = 1.18

The required probability is:

P(X ≤ 10) = Pr (z ≤ 1.18)

= 0.8809

3 (w +6 ) = 3w + 18O Distributive Property
O
Commutative Property
O Associative Property
O Transitive Property

Answers

Answer:

Distributive Property (first option)

Step-by-step explanation:

We apply here the distributive property in order to eliminate parenthesis. That is to "distribute" the factor that is outside (3) into the two terms inside the parenthesis, which cannot be combined because they are not like terms.

Use a Venn diagram to answer the question. A survey of 180 families showed that 67 had a dog; 52 had a cat; 22 had a dog and a cat; 70 had neither a cat nor a dog, and in addition did not have a parakeet; 4 had a cat, a dog, and a parakeet. How many had a parakeet only?

Answers

Answer:

There are 13 families had a parakeet only

Step-by-step explanation:

* Lets explain the problem

- There are 180 families

- 67 families had a dog

- 52 families had a cat

- 22 families had a dog and a cat

- 70 had neither a cat nor a dog, and in addition did not have a

parakeet

- 4 had a cat, a dog, and a parakeet (4 is a part of 22 and 22 is a part

of 67 and 520

* We will explain the Venn-diagram

- A rectangle represent the total of the families

- Three intersected circles:

C represented the cat

D represented the dog

P represented the parakeet

- The common part of the three circle had 4 families

- The common part between the circle of the cat and the circle of the

dog only had 22 - 4 = 18 families

- The common part between the circle of the dog and the circle of the

parakeet only had a families

- The common part between the circle of the cat and the circle of the

parakeet only had b families

- The non-intersected part of the circle of the dog had 67 - 22 - a =

45 - a families

had dogs only

- The non-intersected part of the circle of the cat had 52 - 22 - b =

30 - b families

had cats only

- The non-intersected part of the circle of the parakeet had c families

had parakeets only

- The part out side the circles and inside the triangle has 70 families

- Look to the attached graph for more under stand

∵ The total of the families is 180

∴ The sum of all steps above is 180

∴ 45 - a + 18 + 4 + 30 - b + b + c + a + 70 = 180 ⇒ simplify

- (-a) will cancel (a) and (-b) will cancel (b)

∴ (45 + 18 + 4 + 30 + 70) + (-a + a) + (-b + b) + c = 180

∴ 167 + c = 180 ⇒ subtract 167 from both sides

∴ c = 180 - 167 = 13 families

* There are 13 families had a parakeet only

H=10 cm
r= 7cm
calculate the volume of the cylinder

Answers

Answer:

V = 1538.6 cm^3

Step-by-step explanation:

The volume of a cylinder is given by

V = pi r^2h where r is the radius and h is the height

V = pi (7)^2 10

V = 3.14 (49)10

V = 1538.6 cm^3

Answer:

≈ 1539.3804

Step-by-step explanation:

V= πr²h =π*7²*10

Use the given degree of confidence and sample data to construct a confidence interval for the population mean mu μ. Assume that the population has a normal distribution. A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 185 milligrams with s equals =17.6 milligrams. Construct a 95% confidence interval for the true mean cholesterol content of all such eggs A. 175.9 mg less than < mu μ less than <194.1 mg
B. 173.9 mg less than < mu μ less than <196.1 mg
C. 173.8 mg less than < mu μ less than <196.2 mg
D. 173.7 mg less than < mu μ less than <196.3 mg

Answers

Answer:

option (C) 173.8 mg less than < mu μ less than <196.2 mg

Step-by-step explanation:

Data provided ;

number of sample, n = 12

Mean = 185 milligram

standard deviation, s = 17.6 milligrams

confidence level = 95%

α = 0.05 [for 95% confidence level]

df = n - 1 = 12 - 1 = 11

Now,

Confidence interval = Mean ± E

here,

E is the margin of error = $t_(\alpha/2, df)(s)/(√(n))$

also,

$t_(\alpha/2, df)$

= $t_(0.05/2, (11))$

= 2.201 [ from standard t value table]

Thus,

E = $2.201*(17.6)/(√(12))$

E = 11.182 milligrams ≈ 11.2 milligrams

Therefore,

Confidence interval:

Mean - E < μ < Mean + E

185 - 11.2 < μ < 185 + 11.2

173.8 < μ < 196.2

Hence,

the correct answer is option (C) 173.8 mg less than < mu μ less than <196.2 mg

Final answer:

To construct a confidenceinterval for the population mean cholesterol content of all chicken eggs with a 95% confidence level, we use the sample mean, standard deviation, and sample size to calculate the margin of error. The confidence interval is then constructed by subtracting the margin of error from the sample mean and adding it to the sample mean.

Explanation:

To construct a confidenceinterval for the population mean cholesterol content of all chicken eggs, we first need to find the margin of error. The margin of error depends on the samplemean, standard deviation, sample size, and the desired level of confidence. In this case, we have a sample mean of 185 mg, a standard deviation of 17.6 mg, and a sample size of 12. Since we want a 95% confidence interval, we use a z-score of 1.96. The margin of error is then calculated as 1.96 * (17.6/sqrt(12)), which is approximately 9.61 mg. We can then construct the confidenceinterval by subtracting the margin of error from the sample mean and adding it to the sample mean. Therefore, the 95% confidence interval for the true mean cholesterol content of all such eggs is 175.9 mg to 194.1 mg.