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3 years ago

What are acceptable ways to represent the statement "x is a distance of five from 0." Choose ALL that apply.

Mathematics

2 answers:

Sergio039 [100]3 years ago

6 0

A,b,c d is wrong I just did that question

Ahat [919]3 years ago

5 0

Hello! You should select a, b, and c.
Think of it like on a number line, x could be 5, or -5 and still be the same distance from 0. Absolute vales are the same way, so if the answer is 5 or -5, its right! It cant be the last one because x, must only be 5 or -5. Hope that helps!

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An item is regularly priced at 85$. It is on sale for 80% off the regular price. How much is discounted from the regular price?

laila [671]

$68 is discounted from the regular price.

To find this, you first divide 80% by 100 to get 0.8 (the decimal form of the percentage). Then, simply multiply 0.8 by 85 (or vise verse) to get $68. Since the question is asking for how much is discounted from the regular price, the final answer would be $68.

I hope this helps!

4 0

3 years ago

Suppose an angle α is drawn in standard position. In each case, use the information to determine what quadrant α is in.

cupoosta [38]

Since sin(α) > 0 and cos(α) > 0
so α is IN THE FIRST QUADRANT (I)

4 0

3 years ago

How do you know if a vertex is a minimum or maximum

lara [203]

If the parabola looks like an “n,” your vertex will be a maximum. If the parabola looks like a “u,” the vertex will be a minimum.

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

Each of n specimens is to be weighed twice on the same scale. Let Xi and Yi denote the two observed weights for the ith specimen

Setler [38]

Answer:

attached below

Step-by-step explanation:

Attached below is a detailed solution to your problem above

a) show that the maximum likelihood estimator of Ï2 is Xi and Yi.

attached below is the detailed solution

8 0

3 years ago