(Statistics) Can someone help explain how to do this

Solve the inequality |* +31 < |2x + 1. |x+3|<|2x+1|

Kisachek [45]

Answer:

Step-by-step explanation:

Here are the steps to follow when solving absolute value inequalities:

Isolate the absolute value expression on the left side of the inequality.

If the number on the other side of the inequality sign is negative, your equation either has no solution or all real numbers as solutions.

If your problem has a greater than sign (your problem now says that an absolute value is greater than a number), then set up an "or" compound inequality that looks like this:

(quantity inside absolute value) < -(number on other side)

OR

(quantity inside absolute value) > (number on other side)

The same setup is used for a ³ sign.

If your absolute value is less than a number, then set up a three-part compound inequality that looks like this:

-(number on other side) < (quantity inside absolute value) < (number on other side)

The same setup is used for a £ sign

7 0

3 years ago

The Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are

almond37 [142]

Answer:

For x = 0, P(x = 0) = 0.35

For x = 1, P(x = 1) = 0.54

For x = 2, P(x = 2) = 0.11

For x = 3, P(x = 3) = 0

Step-by-step explanation:

We are given that the Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are not. Three altimeters are randomly selected, one at a time, without replacement.

Let X = the number that are not correctly calibrated.

Number of altimeters that are correctly calibrated = 6

Number of altimeters that are not correctly calibrated = 2

Total number of altimeters = 6 + 2 = 8

(a) For x = 0: means there are 0 altimeters that are not correctly calibrated.

This means that all three selected altimeters are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 3 altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_3$

So, the required probability = $\frac{^{6}C_3}{^{8}C_3}$

= $\frac{20}{56}$ = 0.35

(b) For x = 1: means there is 1 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 1 is not correctly calibrated and 2 are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 2 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_2$

The number of ways of selecting 1 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_1$

So, the required probability = $\frac{^{6}C_2 \times ^{2}C_1 }{^{8}C_3}$

= $\frac{30}{56}$ = 0.54

(c) For x = 2: means there is 2 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 2 are not correctly calibrated and 1 is correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 1 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_1$

The number of ways of selecting 2 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_2$

So, the required probability = $\frac{^{6}C_1 \times ^{2}C_2 }{^{8}C_3}$

= $\frac{6}{56}$ = 0.11

(d) For x = 3: means there is 3 altimeter that is not correctly calibrated.

This case is not possible, so this probability is 0.

6 0

3 years ago

Pls help on #7 I will give brainiest ! 7th grade level

valentina_108 [34]

Answer:

50

Step-by-step explanation:

To solve this problem you multiply 6 by 100 and then divide the total by 12 as follows: (6 x 100) / 12

4 0

2 years ago

Solve each equation write each equation in simplest form 7x/8=21

AveGali [126]

7x / 8 = 21
7x = 8*21
x = 24

8 0

3 years ago

Which equation represents the inverse

STatiana [176]

I'd say A, since there, all the operations are the opposite from the original function.

Hope it helped,

BioTeacher101

4 0

3 years ago