All categories

3 years ago

Solve the equation. Check your answer. If necessary, round to 3 decimal places In x = -2, x=

Mathematics

1 answer:

Nana76 [90]3 years ago

6 0

Answer:

x = 0.135 ( 3 dec. places )

Step-by-step explanation:

Using the law of logarithms

• $log_{b}$ x = n ⇔ x = $b^{n}$

Note that ln x = $log_{e}$ x

given

ln x = - 2 ⇒ x = $e^{-2}$ ≈ 0.135

You might be interested in

TIME SENSITIVE (30 POINTS)

Nonamiya [84]

Answer:

They have different names?

Step-by-step explanation:

brainliest?

6 0

2 years ago

Read 2 more answers

Mumz [18]

martemat 20 e7 72gsys y3823 485 u

8 0

2 years ago

A box of cereal is 13 ounces. About how many grams is it?

DaniilM [7]

1 ounce = 28.3495 grams

13 x 28.3495 = 368.544 grams

round answer as needed

5 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.