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3 years ago

HELP PLEASE!!! 30POINTS

Mathematics

2 answers:

goblinko [34]3 years ago

7 0

Here are table and graph on the same page.

_____

The parent absolute value function is expanded vertically by a factor of 2, then translated right 4 units and up 3 units.

Helga [31]3 years ago

3 0

Answer:

$\boxed{\mathrm{see \: below}}$

Step-by-step explanation:

Domain is the x values.

x = 2, 3, 4, 5, 6

Find the outputs (y) for the inputs (x).

y = 7, 5, 3, 5, 7

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PLEASE HELPP!! RIGHT ANSWERS WILL GET BRAINLIEST!!!

Alecsey [184]

Answer:

Step-by-step explanation:

8 0

3 years ago

How do you do this question?

Luda [366]

Answer:

∫ₑ°° 1 / (x (ln x)¹⁰) dx

∫₁°° x⁻³ dx

Step-by-step explanation:

A p-series 1 / xᵖ converges if p > 1.

∫ₑ°° 1 / (x (ln x)¹⁰) dx

If u = ln x, then du = 1/x dx.

When x = e, u = 1. When x = ∞, u = ∞.

= ∫₁°° 1 / (u¹⁰) du

p = 10, converges

∫₁₀°° x^(-⅔) dx

= ∫₁₀°° 1 / (x^⅔) dx

p = ⅔, diverges

∫₁°° 2 / x^0.5 dx

= 2 ∫₁°° 1 / x^0.5 dx

p = 0.5, diverges

∫₁°° x⁻³ dx

= ∫₁°° 1 / x³ dx

p = 3, converges

∫₂°° 1/(3x) dx

= ⅓ ∫₂°° 1/x dx

p = 1, diverges

8 0

3 years ago

Please help me ! I don’t know !

goblinko [34]

Answer:

Step-by-step explanation:

this will endure the bitextular number course and follow the index pattern once you do that follow the question to get 5

7 0

3 years ago

Read 2 more answers

Find limit as x approaches 1 for x-1/ x^2-3x+2

zysi [14]

I'm going to assume that you meant the following:

(x-1)

---------------

x^2-3x+2

Factoring the denominator, we get:

(x-1)

------------

(x-1)(x-2)

Notice that we have x-1 in both the numerator and the denominator. This factor cancels. HOWEVER: x can still not equal 1, which would lead to div. by zero. With the factor x-1 cancelled, we have 1 / (x-2). As x approaches 1 from either side, y approaches 1 / (1-2), or 1 / (-1), or -1. Thus, the limit exists and is -1.

8 0

3 years ago

A half-century ago, the mean height of women in a particular country in their 20s was 64.7 inches. Assume that the heights of

Ainat [17]

Answer:

99.5% of all samples of 21 of today's women in their 20's have mean heights of at least 65.86 inches.

Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of today's women in their 20's have been taken.

Let $\bar X$ = sample mean heights

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean height of women = 64.7 inches

$\sigma$ = standard deviation = 2.07 inches

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the sample of 21 of today's women in their 20's have mean heights of at least 65.86 inches is given by = P( $\bar X$ $\geq$ 65.86 inches)

P( $\bar X$ $\geq$ 65.86 inches) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{65.86-64.7}{\frac{2.07}{\sqrt{21} } }$ ) = P(Z $\geq$ -2.57) = P(Z $\leq$ 2.57)

= 0.99492 or 99.5%

The above probability is calculated by looking at the value of x = 2.57 in the z table which has an area of 0.99492.

Therefore, 99.5% of all samples of 21 of today's women in their 20's have mean heights of at least 65.86 inches.

3 0

3 years ago

﻿HELP PLEASE!!! 30POINTS

HELP PLEASE!!! 30POINTS