What equation is graphed in this figure? y = -x + 2 y = -x + 3/2 y = 3/2x + 1 y = -3/2x + 1

We already know that Emma is 12, and if Luke is twice her age, then that means we need to multiply 12 by 2, which equals 24. If Kendra is one half of Luke's age, which is 24, then we must divide 24 by 2, which equals 12.
In conclusion, Kendra is 12 years old.

5 0

3 years ago

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A theater has a discount pass that cost nine dollars with the past admission is $6.50 per movie. mike tyrell and juanita all bou

tia_tia [17]

Answer:

They saw 24 movies.

Step-by-step explanation:

The question is incomplete so the complete question is below:

A theater has a discount pass that cost nine dollars with the pass, admission is $6.50 per movie. Mike, Tyrell and Juanita all bought a pass and went to see movies together they spent a total of $183. Write and solve an equation to find out how many movies they saw.

Now, to find the number of movies they saw.

Let the number of movies they saw be $x.$

Discount pass = $9.

With the pass admission per movie is $6.50.

Total number of people = 3.

Total money spent = $183.

So, to get the number of movies they saw we write an equation:

$3(9)+6.50(x)=183$

Now, to solve the equation:

$3(9)+6.50(x)=183$

$27+6.50x=183$

Subtracting on both sides by 27 we get:

$6.50x=156$

Dividing both sides by 6.50 we get:

$x=24.$

Therefore, they saw 24 movies.

4 0

3 years ago

A gondola (cable car) at a ski area holds 50 people. Its maximum safe load is 10000 pounds. A population of skiers has a distrib

fenix001 [56]

Answer: 0.03855

Step-by-step explanation:

Given :A population of skiers has a distribution of weights with mean 190 pounds and standard deviation 40 pounds.

Its maximum safe load is 10000 pounds.

Let X denotes the weight of 50 people.

As per given ,

Population mean weight of 50 people = $\mu=50\times190=9500\text{ pounds}$

Standard deviation of 50 people $=\sigma=40\sqrt{50}=40(7.07106781187)=282.84$

Then , the probability its maximum safe load will be exceeded =

$P(X>10000)=P(\dfrac{X-\mu}{\sigma}>\dfrac{10000-9500}{282.84})\\\\=P(z>1.7671-8)\\\\=1-P(z\leq1.7678)\ \ \ \ [\because\ P(Z>z)=P(Z\leq z)]\\\\=1-0.96145\ \ \ [\text{ By p-value of table}]\\\\=0.03855$

Thus , the probability its maximum safe load will be exceeded = 0.03855

3 0

3 years ago