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Choose the word phrase (greater than, less than, or equal to) to make the statements true.

d1i1m1o1n [39]

Answer:

Five pencils and two marches are worth greater than thirty paper clips.

Step-by-step explanation:

Let 'x' be paper clip.

Let 'y' be a match

and let 'z' be a pencil.

Now, let's write each statement as an equation:

Each paper clip can be traded for three matches:

x = 3y

Each pencil can be traded for six paper clips:

z = 6x

Now, five pencils equals 30 paper clips and two matches equals two thirds papel clips.

Therefore, we can say that Five pencils and two matches equals 30.66 matches. Therefore, five pencils and two marches are worth greater than thirty paper clips.

3 0

2 years ago

Read 2 more answers

Tyresha bought an outfit for 25% off. Before the discount , the items totalled $64. how much did she pay with the discount and 6

Svetach [21]

ANSWER:

50.88

STEP BY STEP

First you would do 64 x .25 which is 16

Then you would take the 16 away from 64 so it would be 48

Then you have to add the tax so it would 48 x .06 = 2.88

Finally you would add the 2.88 to 48 which is 50 .88

3 0

2 years ago

What is the slope of ?

Tamiku [17]

(d-0)/(c-e) - slope CE

6 0

2 years ago

The diagram shows corresponding lengths in two similar figures. Find the area of the smaller figure.

Alika [10]

36cm^2 is your answer

5 0

2 years ago

Consider random samples of size 86 drawn from population A with proportion 0.43 and random samples of size 60 drawn from populat

Elan Coil [88]

Answer:

a) The standard error is s = 0.071.

b) Yes, as both sample sizes are above 30.

Step-by-step explanation:

To solve this question, we need to understand the central limit theorem and subtraction of normal variables.

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 $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Samples of size 86 drawn from population A with proportion 0.43

This means that $n = 86, p = 0.43$ . So:

$s_A = \sqrt{\frac{0.43*0.57}{86}} = 0.0534$

Samples of size 60 drawn from population B with proportion 0.15:

This means that $n = 60, p = 0.15$ . So:

$s_B = \sqrt{\frac{0.15*0.85}{60}} = 0.0461$

(a) Find the standard error of the distribution of differences in sample proportions A - B Round your answer for the standard error to three decimal places. Standard error=

This is:

$s = \sqrt{s_A^2 + s_B^2}$

$s = \sqrt{(0.0534)^2 + (0.0461)^2}$

$s = 0.071$

The standard error is s = 0.071.

(b) Are the sample sizes large enough for the Central Limit Theorem. Yes or No?

Yes, as both sample sizes are above 30.

5 0

2 years ago