All categories

3 years ago

Question 4

Mathematics

1 answer:

mel-nik [20]3 years ago

8 0

Answer:

choice 4) 33.5 in³

Step-by-step explanation:

r = 4/2

V = 4/3πr³ = 4/3(3.14)(2³) = 33.5 in²

You might be interested in

140 fluid ounces increased by 45%

Radda [10]

$100\%\ increased\ by\ 45\%=100\%+45\%=145\%\\\\method\ \#1\\\\p\%=\frac{p}{100}\to145\%=\frac{145}{100}=\frac{145:5}{100:5}=\frac{29}{20}\\\\145\%\ of\ 140=\frac{29}{20}\cdot140=29\cdot7=\boxed{203\ (ounces)}\leftarrow answer$

$method\ \#2\\\\\begin{array}{ccc}140&-&100\%\\x&-&145\%\end{array}\ \ \ cross\ multiply\\\\\\100\cdot x=140\cdot145\ \ \ \ |divide\ both\ sides\ by\ 100\\\\x=\frac{140\cdot145}{100}\\\\\boxed{x=203\ (ounces)}$

4 0

4 years ago

The screen on a computer monitor has a length of 15 inches and a width of 12 inches. What is the length of the diagonal to the n

Mekhanik [1.2K]

Answer:

19.2

Step-by-step explanation:

You use the pyth.

15^2+12^2=369

Then find the square root of 392 which is 19.2

I hope this helps! have a good day :)

5 0

3 years ago

Which expression shows a sum of three terms.

vladimir1956 [14]

Answer: B) 3+y+3

This can be simplified to y+6, but the current un-simplified expression has 3 terms.

======================================

Explanation:

Terms are separated by a plus sign. If you had something like 10x-5y, then you would write that as 10x+(-5y) showing that 10x and -5y are the two terms.

Choices A and C, xy and 6y respectively, have one term each. They are considered monomials. Mono = one, nomial = name.

Choice D is the product of the constant 3 and the binomial y+3. Binomials have two terms.

Only choice B has three terms, though we can simplify it down to two terms. I have a feeling your teacher doesn't want you to simplify it.

5 0

3 years ago

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol

bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.

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.

25% of the incoming calls involve fax messages

This means that $p = 0.25$

25 incoming calls.

This means that $n = 25$

a. What is the probability that at most 4 of the calls involve a fax message?

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$ .

In which

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

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214$

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$ . Then

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904$

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

$P(X \leq 4) + P(X > 4) = 1$

From a), $P(X \leq 4) = 0.214)$

Then

$P(X > 4) = 1 - 0.214 = 0.786$

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

8 0

3 years ago

Determine the sampling error if the grade point averages for 10 randomly selected students from a class of 125 students has a me

zaharov [31]

The sampling error if the grade point averages for 10 randomly selected students from a class of 125 students has a mean of x =2. is A. -0.7.

Given:

Random number=10

mean of x=2

population=125

mean of p=2.7

Using this formula

Sampling error=x− μ

Where:

x= sample mean=2

μ=population mean=2.7

Let plug in the formula

Sampling error=2-2.7

Sampling error=-0.7

Inconclusion the sampling error is A.-0.7.

Learn more here:

brainly.com/question/13929265

3 0

3 years ago