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3 years ago

Where is the answer to the expression 4 − 8 located on a horizontal number line? 4 units to the left of 4

1 answer:

4 0

4 - 8 = -4

A negative number is alwats at the left in a horizontal line.

Let's try to rapresent it.

-4 -3 -2 -1 0 +1 +2 +3 +4

So, you count from +4 to -4

Your answer is "8 units to the left of 4"

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HELP! the student council at woodlands middle school is planning a field trip. they polled randomly selected teachers and studen

gregori [183]

Answer:

Step-by-step explanation:

because on the chart is says 19% under burgers for teachers.

Tell me if it is right.

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2 years ago

Suppose that the populations of the United States and China both increase by 12 million people in one year. What would be the re

Jlenok [28]

Question:

Suppose that the populations of the United States and China both increase by 12 million people in one year. What would be the relative change for each country? Answer to the nearest tenth of a percent (1 decimal place). Make sure that you fill in each answer area before checking "How Did I Do?".

If you assume that the U.S. population is about 300 million, an increase of 12 million would result in a relative change of Number %.

If you assume that China has about 1 billion people, an increase of 12 million would result in a relative change of Number %.

Answer:

The relative change of the United States and China is 4% and 1.2%, respectively.

Step-by-step explanation:

Given

Represent population with p. So, we have:

United States

$\triangle p = 12\ million$

$p_1 =300\ million$

China

$\triangle p = 12\ million$

$p_1 =1\ billion$

Required

Determine the relative change for each country

Relative change, C is calculated as:

$C = \frac{\triangle p}{p} * 100\%$

For United States

$C = \frac{12\ million}{300\ million} * 100\%$

$C = \frac{12}{300} * 100\%$

$C = \frac{12 * 100}{300}\%$

$C = \frac{1200}{300}\%$

$C = 4\%$

For China

$C = \frac{12\ million}{1\ billion} * 100\%$

$C = \frac{12}{1000} * 100\%$

$C = \frac{12* 100}{1000} \%$

$C = \frac{1200}{1000} \%$

$C = 1.2\%$

The relative change of the United States and China is 4% and 1.2%, respectively.

3 0

2 years ago

PLEASE HELP DUE SOON MORE POINTS IF CORRECT!!

SVETLANKA909090 [29]

Answer:

y=4x+35

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

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

3 years ago

Use the ratio go solve the percent problem

Drupady [299]

Answer
40%
32 is what percent of 80? =40.

5 0

3 years ago