Given: ∠BCD ≅ ∠EDC and ∠BDC ≅ ∠ECD Prove: Δ BCD ≅ Δ EDC

aniked [119]

yes!! glad that you support this! keep your head up and keep on smiling! <3

5 0

3 years ago

Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati

Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c. D. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X $\sim$ N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

$P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})$

$P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})$

$P(X < 76) = P(Z< \dfrac{3}{12.5})$

$P(X < 76) = P(Z< 0.24)$

From the standard normal distribution tables,

$P(X < 76) = 0.5948$

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

$P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})$

$P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})$

$P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})$

$P( \overline X < 76) = P(Z< 1.2)$

From the standard normal distribution tables,

$P(\overline X < 76) = 0.8849$

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

In order to determine the probability in part (b); the normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

5 0

2 years ago

(Four-fifths + three-fifths) + 3.5 times 5

defon

Answer:

18.9

Step-by-step explanation:

First, lets solve for the fractions. You can divide the fractions from the top, to their bottom (4 divided by 5, 3 divided by 5), in this case, you get 1.4. Now you have to rewrite the problem: 1.4+3.5x5. You solve it, and get 18.9.

7 0

3 years ago

Read 2 more answers

Show work, please someone help me, i'm confused still on this concept.

BigorU [14]

Answer: choice C) -15x^4y

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Explanation:

The coefficients are -3 and 5. They are the numbers to the left of the variable terms
Multiply the coefficients to get -3*5 = -15. So -15 is the coefficient in the answer

Multiply the x terms to get x^3 times x = x^(3+1) = x^4. Notice the exponents are being added

Do the same for the y terms as well: y^2 times y^(-1) = y^(2+(-1)) = y^(2-1) = y^1 = y

So we have a final coefficient of -15, the x terms simplify to x^4 and the y terms simplify to just y

Put this all together and we end up with -15x^4y which is what choice C is showing

3 0

3 years ago

If candies are sold at 3 pcs for php.2.00.how many candies can melai get if she has php20.00?

Marrrta [24]

If you’re saying that 3 pieces costs 2.00 then the answer is 30. 2x10 is the $20.00 so 3x10 = 30 she can buy

8 0

3 years ago