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3 years ago

Which graph represents the function f(x)=|x+2|+1

Mathematics

2 answers:

inessss [21]3 years ago

7 0

This graph is a "v" shape with the vertex at (-2,1) and points (0,3) and (-5,4)

This graph is in quadrant 2 and 1

Snezhnost [94]3 years ago

4 0

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

$f\left(x\right)=\left|x+2\right|+1$

Using a graphing tool

The graph is a "V" shape

The vertex is the point $(-2,1)$

The y-intercept is the point $(0,3)$

see the attached figure

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BRAINLIEST FOR THE RIGHT ANSWERRR If useful, points are (1,0) (0,-4)

Lapatulllka [165]

Answer:

x-inercept = 1 y-intercept = -4

Step-by-step explanation:

3 0

2 years ago

Find the value of y.

Leno4ka [110]

Answer:

$\sqrt{55}$

Step-by-step explanation:

Based off of what we do know, if we can find length DB we can use the pythagorous theorem (I will be assigning the variable z)

$x^{2} =36+z^{2} \\\\y^{2}=25+z^2\\\\x^{2} +y^2=121\\\\$

First equation is for triangle ABD

Second is for BCD

And the last one is for ACD

Substitute $x^{2}$ and $y^2$

$36+z^2+25+z^2=121\\\\61+2z^2=121\\\\2z^2=60\\\\z^2=30\\\\z=\sqrt{30}$

As discovered before

$y^2=25+z^2\\\\y^2=25+30\\\\y^2=55\\\\y=\sqrt{55}$

8 0

3 years ago

Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72

Lisa [10]

Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

To solve these questions, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that $\mu = 38.72, \sigma = 3.17$

Sample of 10:

This means that $n = 10, s = \frac{3.17}{\sqrt{10}}$

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}$

$Z = 1.28$

$Z = 1.28$ has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

$\mu = 266, \sigma = 16$

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{260 - 266}{16}$

$Z = -0.375$

$Z = -0.375$ has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now $n = 20$ , so:

$Z = \frac{X - \mu}{s}$

$Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}$

$Z = -1.68$

$Z = -1.68$ has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now $n = 50$ , so:

$Z = \frac{X - \mu}{s}$

$Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}$

$Z = -2.65$

$Z = -2.65$ has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that $n = 15$ . This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

$Z = \frac{X - \mu}{s}$

$Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}$

$Z = 2.42$

$Z = 2.42$ has a p-value of 0.9922.

X = 256

$Z = \frac{X - \mu}{s}$

$Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}$

$Z = -2.42$

$Z = -2.42$ has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

8 0

3 years ago

Mabel bought b bags of beans. There are 15 beans in each bag. Write an expression that shows how many beans Mabel bought.

denpristay [2]

Answer:

y= x+15

Step-by-step explanation:

I think that will be your equation

8 0

3 years ago

Help I need answers

natali 33 [55]

We can easily work out the meter reading in October. It comes out to be 32347 + 972. So the reading in October must be around 33319.

SO the answer to second question is really easy to figure out too. As we know that one unit costs 14 p we can simply multiply the cost with the number of units consumed.

So for 972 units if we multiply by 14 we get the cost to be around somewhere 13608 p.

6 0

3 years ago