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4 years ago

Suppose △FLX≅△ZAQ.

Mathematics

1 answer:

Oksi-84 [34.3K]4 years ago

3 0

When two triangles are congruent, the letters correspond with each other.

Because △FLX≅△ZAQ, so F corresponds with Z, L with A, and X with Q.

So as long as they're in the same place within the term, the term would be correct.

So △XFL≅△QZA, △LFX≅△AZQ, and △FXL≅△ZQA are correct.

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Hi guys i need help with answering this question, it says put the following numbers in order greatest to least

mr Goodwill [35]

Answer:

12.47, 12.075, 11 1/4, 11, -12, -12.004, -12.01

8 0

3 years ago

Read 2 more answers

the measures of an exterior angle and the adjacent interior angle add p to 360 because they form a linear plane?

inn [45]

Sadly, I can't see the picture you're looking at as you make that statement.
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6 0

4 years ago

Teen obesity:

Law Incorporation [45]

Answer:

95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].

Step-by-step explanation:

We are given that 15% of a random sample of 300 U.S. public high school students were obese.

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample % of U.S. public high school students who were obese = 15%

n = sample of U.S. public high school students = 300

p = population percentage of all U.S. public high school students

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.15-1.96 \times {\sqrt{\frac{0.15(1-0.15)}{300} } }$ , $0.15+1.96 \times {\sqrt{\frac{0.15(1-0.15)}{300} } }$ ]

= [0.110 , 0.190]

Therefore, 95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].

6 0

4 years ago

What is the absolute value of y+3=2

Artyom0805 [142]

Answer:

y = -1

Step-by-step explanation:

8 0

3 years ago

A DVD rental store wants to know what proportion of its customers are under age 21. A simple random sample of 400 customers was

Volgvan

Answer:

The sample proportion $\hat{p}$ is within 0.03 of the true proportion of customers who are under age 21 is 0.803

Step-by-step explanation:

Total no. of customers = n = 400

We are given that the true population proportion of customers under age 21 is 0.68.

So, p =0.68

q=1-p=1-0.68=0.32

Standard deviation = $\sqrt{\frac{pq}{n}}=\sqrt{\frac{0.68 \times 0.32}{400}}= 0.023$

We are supposed to find the probability that the sample proportion $\hat{p}$ is within 0.03 of the true proportion of customers who are under age 21 that is , what is the probability that $\hat{p}$ is between 0.68 - 0.03 and 0.68+ 0.03

$P(0.65 < X < 0.71) = P((\frac{0.65-0.68}{0.0233}) < Z < (\frac{0.71-0.68}{0.0233}))$

Using Z table

$= P(-1.2875 < Z < 1.2875)= 0.803$

Hence the sample proportion $\hat{p}$ is within 0.03 of the true proportion of customers who are under age 21 is 0.803

5 0

4 years ago