If ∠2 and ∠3 are supplementary angles, then ∠2 and ∠3 form a linear pair true or false

Nataly [62]

The answer should be 60 (sorry if it’s wrong)

5 0

3 years ago

Justin and Eric came home from school on Friday and found a freshly baked cherry pie. Justin ate 2/3 of a pie and Eric ate 1/4 o

Verdich [7]

Answer:

1/12

Step-by-step explanation:

3 0

2 years ago

Pam jogged up the hill at 6 km/h and then jog down back down the hill at 10 km an hour how many kilometers did you travel at all

atroni [7]

Answer: you travelled 10 km

Step-by-step explanation:

Pam jogged up the hill at 6 km/h.

Let x represent the number of kilometers that you travelled by jogging up the hill or returning back.

Time = Distance/speed

Pam jogged up the hill at 6 km/h.

Time spent in jogging up the hill would be

x /6

She then jog back down the hill at 10 km an hour. Time spent in jogging down the hill would be

x /10

If the total jogging tie is 1 hour 20 minutes, converting to hours, it becomes 1 + 20/60 = 1.33

Therefore

x /6 + x/10 = 1.33

10x + 6x = 80

16x = 80

x = 80/16 = 5

Total distance travelled up and down the hill would b

5 × 2 = 10 km

4 0

4 years ago

What is 696 divided by 12 and show your work please

Vinil7 [7]

Answer:

696/12

696/12 math problems division = 58. Therefore 58 to 2 decimal places= 58

696/12 divided by 2 » (696/12) ÷ 2 » 58 ÷ 2 = 29

Step-by-step explanation:

The answer is 58 mark me brainliest

4 0

3 years ago

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

frosja888 [35]

Answer:

a) 0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

b) 0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday

c) 0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday

d) 0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday

Step-by-step explanation:

We solve this question using the normal approximation to the binomial distribution.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

Sample of 723, 3.7% will live past their 90th birthday.

This means that $n = 723, p = 0.037$ .

So for the approximation, we will have:

$\mu = E(X) = np = 723*0.037 = 26.751$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{723*0.037*0.963} = 5.08$

(a) 15 or more will live beyond their 90th birthday

This is, using continuity correction, $P(X \geq 15 - 0.5) = P(X \geq 14.5)$ , which is 1 subtracted by the pvalue of Z when X = 14.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{14.5 - 26.751}{5.08}$

$Z = -2.41$

$Z = -2.41$ has a pvalue of 0.0080

1 - 0.0080 = 0.9920

0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

(b) 30 or more will live beyond their 90th birthday

This is, using continuity correction, $P(X \geq 30 - 0.5) = P(X \geq 29.5)$ , which is 1 subtracted by the pvalue of Z when X = 29.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{29.5 - 26.751}{5.08}$

$Z = 0.54$

$Z = 0.54$ has a pvalue of 0.7054

1 - 0.7054 = 0.2946

0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday

(c) between 25 and 35 will live beyond their 90th birthday

This is, using continuity correction, $P(25 - 0.5 \leq X \leq 35 + 0.5) = P(X 24.5 \leq X \leq 35.5)$ , which is the pvalue of Z when X = 35.5 subtracted by the pvalue of Z when X = 24.5. So