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valina [46]
4 years ago
15

Few math questions TIME IS RUNNING OUT!! PLEASE HELP!!!!

Mathematics
1 answer:
olya-2409 [2.1K]4 years ago
3 0
1.

|5+6x|\ \textless \ 13\\\\-13\ \textless \ 5+6x\ \textless \ 13

so

\begin{cases}5+6x\ \textless \ 13\\5+6x\ \textgreater \ -13\end{cases}

Answers B and D.

2.

|x+9|\geq11\\\\x+9\ \textgreater \ 11\qquad\vee\qquad x+9\leq-11\\\\
\boxed{x\geq2\qquad\vee\qquad x\leq-20}

Answer D.

3.

The same as point 2. Answer D.

4.

Answer C.

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What is the solution to the inequality |2n+5|>1?
Yuliya22 [10]
ANSWER

n <  - 3 \: or \: n>  - 2



EXPLANATION



The given inequality is,

|2n + 5|  \:  >  \: 1


By the definition of absolute value,



- (2n + 5)  \:  >  \: 1 \: or \: (2n + 5) \:  >  \: 1



We divide through by negative 1, in the first part of the inequality and reverse the sign to get,

2n + 5 \:   <   \:  - 1 \: or \: (2n + 5) \:  >  \: 1

We simplify now to get,

2n   \:   <   \:  - 1 - 5 \: or \: 2n  \:  >  \: 1 - 5


2n   \:   <   \:  - 6 \: or \: 2n  \:  >  \:  - 4


Divide through by 2 to obtain,

n   \:   <   \:  - 3 \: or \: n  \:  >  \:  - 2


4 0
4 years ago
Read 2 more answers
The population mean annual salary for environmental compliance specialists is about ​$63 comma 500. A random sample of 31 specia
Nat2105 [25]

Answer:

0.0035289

Step-by-step explanation:

From the question;

mean annual salary = $63,500

n = sample size = 31

Standard deviation = $6,200

Firstly, we calculate the z-score of $60,500

Mathematically;

z-score = x-mean/SD/√n = (60500-63500)/6200/√(31) = -2.6941

So we want to find the probability that P(z < -2.6941)

We can get this from the standard normal table

P( z < -2.6941) = 0.0035289

3 0
3 years ago
Im really confused i hope this doesnt take to much time please help it would me so much to me.
katrin [286]
Yes it is A I got it right
6 0
3 years ago
The brand manager for a brand toothpaste must plan a campaign designed to increase brand recognition. He wants to first determin
bulgar [2K]

We need to find out how many adults must the brand manager survey in order to be 90% confident that his estimate is within five percentage points of the true population percentage.

From the given data we know that our confidence level is 90%. From Standard Normal Table we know that the critical level at 90% confidence level is 1.645. In other words, Z_{Critical}=1.645.

We also know that E=5% or E=0.05

Also, since, \hat{p} is not given, we will assume that \hat{p}=0.5. This is because, the formula that we use will have \hat{p}(1-\hat{p}) in the expression and that will be maximum only when \hat{p}=0.5. (For any other value of \hat{p}, we will get a value less than 0.25. For example if, \hat{p} is 0.4, then 1-\hat{p}=0.6 and thus, \hat{p}(1-\hat{p})=0.24.).

We will now use the formula

n=(\frac{Z_{Critical}}{E})^2\hat{p}(1-\hat{p})

We will now substitute all the data that we have and we will get

n=(\frac{1.645}{0.05})^2\times0.5(1-0.5)

n=(32.9)^2\times0.25

n=270.6025

which can approximated to n=271.

So, the brand manager needs a sample size of 271

3 0
3 years ago
What is 6.9 divided by 58.65?
Alecsey [184]

Answer:

8.5

Step-by-step explanation:

3 0
3 years ago
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