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3 years ago

Could someone please help me find the domain and range of the equation?...and B-64 a. Please??

Mathematics

1 answer:

alexandr1967 [171]3 years ago

6 0

<h3>B-61. C=</h3><h3><u>Domain:</u></h3>

0 < x < ∞

x > 0

<h3><u>Range:</u></h3>

y ≥ 500

9 − 3 = 4 − 3

5 − 3 = −3

5 = 0

= 0

Hope this helps you ! :)

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Can segments with lengths of 15, 20, and 36 form a triangle?

strojnjashka [21]

Ok, so remember

the legnth of the longest side must be LESS THAN the sum of the measures of the other 2 sides
if no longest side (becasue it has 2 longest sides or 3 equal sides), then it can form a triangle

so the longest side is 36

36 must be less than the sum of 15 and 20
36<15+20
36<35
false

therfor it cannot form a triangle
try it yourself, you cannot connect them that way

5 0

3 years ago

A man who can row in still water at 5 mph heads directly across a river flowing at 6 mph. At what speed does he drift downstream

Rashid [163]

7.8 mph... Hope it helps

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3 years ago

Read 2 more answers

4x + 1 = 5, find x assuming x is a whole number

levacccp [35]

Answer:

x=1

Step-by-step explanation:

4x+1=5

4x=5-1

4x=4

x=4/4=1

This happens to be a whole number.

4 0

3 years ago

You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard devia

gtnhenbr [62]

Answer:

$n=(\frac{2.58(58.2)}{1})^2 =22546.82 \approx 22547$

So the answer for this case would be n=22547 rounded up to the nearest integer

Step-by-step explanation:

Let's define some notation

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=58.2$ represent the population standard deviation

n represent the sample size

$ME =1$ represent the margin of error desire

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =+1 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. The significance would be $\alpha=0.01$ and the critical value $z_{\alpha/2}=2.58$ , replacing into formula (b) we got:

$n=(\frac{2.58(58.2)}{1})^2 =22546.82 \approx 22547$

So the answer for this case would be n=22547 rounded up to the nearest integer

3 0

3 years ago

The distribution of weights for newborn babies is approximately normally distributed with a mean of 7.4 pounds and a standard de

blsea [12.9K]

Answer:

1. 15.87%

2. 6 pounds and 8.8 pounds.

3. 2.28%

4. 50% of newborn babies weigh more than 7.4 pounds.

5. 84%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 7.4 pounds

Standard Deviation, σ = 0.7 pounds

We are given that the distribution of weights for newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

1.Percent of newborn babies weigh more than 8.1 pounds

P(x > 8.1)

$P( x > 8.1) = P( z > \displaystyle\frac{8.1 - 7.4}{0.7}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 8.1) = 1 - 0.8413 = 0.1587 = 15.87\%$

15.87% of newborn babies weigh more than 8.1 pounds.

2.The middle 95% of newborn babies weight

Empirical Formula:

Almost all the data lies within three standard deviation from the mean for a normally distributed data.
About 68% of data lies within one standard deviation from the mean.
About 95% of data lies within two standard deviations of the mean.
About 99.7% of data lies within three standard deviation of the mean.

Thus, from empirical formula 95% of newborn babies will lie between

$\mu-2\sigma= 7.4-2(0.7) = 6\\\mu+2\sigma= 7.4+2(0.7)=8.8$

95% of newborn babies will lie between 6 pounds and 8.8 pounds.

3. Percent of newborn babies weigh less than 6 pounds

P(x < 6)

$P( x < 6) = P( z > \displaystyle\frac{6 - 7.4}{0.7}) = P(z < -2)$

Calculation the value from standard normal z table, we have,

$P(x < 6) =0.0228 = 2.28\%$

2.28% of newborn babies weigh less than 6 pounds.

4. 50% of newborn babies weigh more than pounds.

The normal distribution is symmetrical about mean. That is the mean value divide the data in exactly two parts.

Thus, approximately 50% of newborn babies weigh more than 7.4 pounds.

5. Percent of newborn babies weigh between 6.7 and 9.5 pounds

$P(6.7 \leq x \leq 9.5)\\\\ = P(\displaystyle\frac{6.7 - 7.4}{0.7} \leq z \leq \displaystyle\frac{9.5-7.4}{0.7})\\\\ = P(-1 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -1)\\= 0.9987 -0.1587= 0.84 = 84\%$

84% of newborn babies weigh between 6.7 and 9.5 pounds.

7 0

4 years ago