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meriva
3 years ago
11

Solve for x under the assumption that x<0. x-5/x<4

Mathematics
1 answer:
AleksandrR [38]3 years ago
3 0
If you would like to solve the inequality x - 5/x < 4, you can do this using the following steps:

<span>x - 5/x < 4
</span>x^2 - 5 < 4x
x^2 - 4x - 5 < 0
(x - 5) * (x + 1) < 0
x < 0
1. x = 5
2. x = -1

The correct result would be x = -1.
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What are ALL the true statements.
ale4655 [162]

Answer:

C, E, F

Step-by-step explanation:

A is false because dilations can not only increase the length of line segments, but also decrease the length of line segments.

B is a true statement.

C is true because when angles undergo dilations, their "size" does not change.

D is false because dilations do not increase the measures of angles.

E is false because a triangle that underwent a dilation becomes similar, not congruent.

F is true because of E's reasoning.

8 0
3 years ago
Read 2 more answers
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
I had to put two screenshot because it would screenshot the whole question but plz help
Tcecarenko [31]

Answer:

i think its D dont fullly take my word tho

Step-by-step explanation:

3 0
3 years ago
In a sample of 42 burritos, we found a sample mean 1.4 lb and assumed that sigma equals.5. In a test of the hypothesis H subscri
Sergio [31]

Answer:

D. 0.9953 (Probability of a Type II error), 0.0047 (Power of the test)

Step-by-step explanation:

Let's first remember that a Type II error is to NOT reject H0 when it is false, and the probability of that occurring is known as β. On the other hand, power refers to the probability of rejecting H0 when it is false, so it can be calculated as 1 - β.

To resolve this we are going to use the Z-statistic:

                                           Z = (X¯ - μ0) / (σ/√n)

where  μ0 = 1.2

            σ = 5

            n = 42

As we can see in part A of the attached image, we have the normal distribution curve representation for this test, and because this is a two-tailed test, we split the significance level of α=0.01 evenly into the two tails, 0.005 in each tail, and if we look for the Z critical value for those values in a standard distribution Z table we will find that that value is 2.576.

Now we need to stablish the equation that will telll us for what values of X¯ will we reject H0.

Reject if:

Z ≤ -2.576                                                          Z ≥ 2.576

We know the equation for the Z-statistic, so we can substitute like follows and resolve.

Reject if:

(X¯ - 1.2) / (5/√42) ≤ -2.576                          (X¯ - 1.2) / (5/√42) ≥ 2.576

X¯ ≤ -0.79                                                         X¯ ≥ 3.19

We have the information that the true population mean is 1.25, so we now for a fact that H0 is false, so with this we can calculate the probability of a Type II error:  P(Do not reject H0 | μ=1.25)

As we can see in part B of the attached image, we can stablish that the type II error will represent the probability of the sample mean (X¯) falling between -0.79 and 3.19 when μ=1.25, and that represents the shaded area. So now we now that we are looking for P(-0.79 < X¯ < 3.19 | μ=1.25).

Because we know the equation of Z, we are going to standardize this as follows:

P ( (-0.79 - 1.25) / (5/√42) < Z <  (3.19 - 1.25) / (5/√42) )

This equals to:

P(-2.64 < Z < 2.51)

If we go and look for the area under the curve for Z positive scores in a normal standart table (part C of attached image), we will find that that area is 0.9940, which represents the probability of a Type II error.

Therefore, the power of the test will be 1-0.9940 = 0.006

If we look at the options of answers we have, there is no option that looks like this results, which means there was a probable redaction error, so we are going to stay with the closest option to these values which is option D.

3 0
3 years ago
Factor the expression: <br> 30 + 100c
ANEK [815]
10(3+10с)
This expression has a common factor of 10
3 0
3 years ago
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