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2 years ago

5

The mean weight of the domesticated house cat is 9.2 pounds with a standard deviation of 3.4 pounds. If a cat has a z-score of -

1.3 for its weight, how much does it weigh? Write your answer as a complete sentence.

1 answer:

ra1l [238]2 years ago

5 0

Answer:

4.78 pounds

Step-by-step explanation:

We solve the above question using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 9.2 pound

σ is the population standard deviation = 3.4 pounds

z = 1.3

We are to find x

Hence,

-1.3 = x - 9.2/3.4

Cross Multiply

-1.3 × 3.4 = x - 9.2

-4.42 = x - 9.2

x = -4.42 + 9.2

x = 4.78 pounds.

Therefore the weight of the cat is 4.78 pounds

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From bottom to top:

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3 years ago

A gardener has a circular flower bed with a radius of 17 feet. If she wants to make a square flower bed with the same area as th

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We know that

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8 0

7 months ago

A,B and C are the vertices of one triangle.

dsp73

Answer:

Step-by-step explanation:

Given: The triangle with coordinate A(4,6), B(2,-2) and C(-2,-4). D is the mid point of AB and E is the mid point of AC.

To prove: DE is parallel to BC.

Construction: Join DE.

Proof: If we prove the basic proportionality theorem that is $\frac{AD}{DB}=\frac{AE}{EC}$ , then it proves that DE is parallel to BC.

Now, Mid Point D has coordinates= $(\frac{4+2}{2},\frac{6-2}{2})=(3,2)$ and Mid Point E has coordinates= $(\frac{4-2}{2},\frac{6-4}{2})=(1,1)$

Now, AD= $\sqrt{(4-3)^{2}+(6-2)^{2}}=\sqrt{17}$

DB= $\sqrt{(3-2)^{2}+(2+2)^{2}}=\sqrt{17}$

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EC= $\sqrt{(1+2)^{2}+(1+4)^{2}}=\sqrt{34}$

Now, $\frac{AD}{DB}=\frac{AE}{EC}$

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4 0

3 years ago

Read 2 more answers

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2 years ago

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Answer:

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Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

Sampling error of 0.03.

This means that $M = 0.03$

99.74% confidence level

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This means that $\pi = 0.25$

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The sample size is n. So

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$0.03 = 3\sqrt{\frac{0.25*0.75}{n}}$

$0.03\sqrt{n} = 3\sqrt{0.25*0.75}$

Simplifying by 0.03 both sides

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The sample size is 1875.

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3 years ago