The difference between a number x and 18 is 22?

At the local recycling center, employees are paid every two weeks. If Jeremy was paid 12,480 last year, how much did he make eac

Alborosie

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4 0

3 years ago

Please help :( Solve the qaudratic equation given below (9x + 13)² = 49(9x + 13)² = 49

NemiM [27]

Answer:

Im pretty sure ur answers would be

x= -32/27 and X= -46/27

Step-by-step explanation:

So srry if im wrong tho

5 0

2 years ago

Can someone please tell me the answer on this.

marishachu [46]

Answer: x = 31

Step-By-Step:

To Find x:

$\left(2x+20\right)+\left(x+10\right)+\left(2x-5\right)=180$ (Angle Sum Property Of Triangles)

(Remove the brackets)

$2x+20+x+10+2x-5=18$

(Group Accordingly)

$2x+x+2x+20+10-5=180$

= $5x+20+10-5=180$

= $5x\:+\:25$ $=180$

= $5x= 180 - 25$

= $5x = 155$

= $x=\frac{155}{5}$

= $31$

Therefore x = 31

So,

Angle A = x + 10 = 31 + 10 = 41

Angle B = 2*x + 20 = 2*31 + 20 = 82

Angle C = 2*x - 5 = 62 - 5 = 57

5 0

2 years ago

5, 10, 12, 4, 6, 11, 13, 5 calculate the median show your work

Zinaida [17]

First you want to organize it from least to greatest: 4, 5, 5, 6, 10, 11, 12, 13

Then look for the one in the middle. But oh wait there are 2 so you need to add 6 and 10 then divide it by 2 to get a median of 8.

A song I learned back then is this I don't know who it is by though: Mode, mode, mode is the most, Average is the mean.Median, median, median, median, The number in between.

8 0

3 years ago

Read 2 more answers

A simple random sample of 90 is drawn from a normally distributed population, and the mean is found to be 138, with a standard d

bagirrra123 [75]

The 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C: [130.10, 143.90]

<h3>
How to find the confidence interval for population mean from large samples (sample size > 30)?</h3>

Suppose that we have:

Sample size n > 30
Sample mean = $\overline{x}$
Sample standard deviation = s
Population standard deviation = $\sigma$
Level of significance = $\alpha$

Then the confidence interval is obtained as

Case 1: Population standard deviation is known

$\overline{x} \pm Z_{\alpha /2}\dfrac{\sigma}{\sqrt{n}}$

Case 2: Population standard deviation is unknown.

$\overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}$

For this case, we're given that:

Sample size n = 90 > 30
Sample mean = $\overline{x}$ = 138
Sample standard deviation = s = 34
Level of significance = $\alpha$ = 100% - confidence = 100% - 90% = 10% = 0.1 (converted percent to decimal).

At this level of significance, the critical value of Z is: $Z_{0.1/2}$ = ±1.645

Thus, we get:

$CI = \overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}\\CI = 138 \pm 1.645\times \dfrac{34}{\sqrt{90}}\\\\CI \approx 138 \pm 5.896\\CI \approx [138 - 5.896, 138 + 5.896]\\CI \approx [132.104, 143.896] \approx [130.10, 143.90]$

Thus, the 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C: [130.10, 143.90]

Learn more about confidence interval for population mean from large samples here:

brainly.com/question/13770164

3 0

2 years ago