Please look at the question in the picture it’s a statistics problem

Kareendeep has $168.50 in her bank account. She buys x shirts for $5.5o each. Write and solve an equation Kareendeep can use to

Vikentia [17]

Answer:

Equation: 5.5x = 168.5

Solution: 30 shirts

Step-by-step explanation:

First, create the equation where x is the number of shirts.

Since each shirt is $5.50, the cost of the shirts will be 5.50x

Set this equal to 168.50, since that is how much money Kareendeep has.

So, the equation will be 5.5x = 168.5

Now, solve this for x, the number of shirts she can buy:

5.5x = 168.5

Divide each side by 5.5

x = 30.6

Since we can only have a whole number of shirts, she can only buy 30 shirts.

So, the answer is 30 shirts.

5 0

3 years ago

One triangle on a graph has a vertical side of 5 and a horizontal side of 11. Another triangle on a graph has a vertical side of

Tamiku [17]

Answer:

Yes.

Step-by-step explanation:

The two triangles are <em>similar</em> because they have:

a common angle (90°)
corresponding sides with lengths in a 3:1 ratio

All you need to do is to slide the smaller triangle into the larger one so that a small angle and one of its sides coincide.

Then the hypotenuses of each triangle will lie along the same line.

6 0

3 years ago

2. A statistics student plans to use a TI-84 Plus calculator on her final exam. From past experience, she estimates that there i

Anarel [89]

Answer:

P(≥1 working) = 0.9936
She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.

Step-by-step explanation:

1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...

... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936

2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.

If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.

This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.

_____

My assessment is that there is significant gain from bringing a backup. (Personally, I might investigate why the probability of failure is so high. I have not had such bad luck with calculators, which makes me wonder if operator error is involved.)

7 0

3 years ago

There are 4 jacks and 13 clubs in a standard, 52-card deck of playing cards. What is the probability that a card picked at rando

ale4655 [162]

Answer:

16/52, or 4/13.

Step-by-step explanation:

First, since we know that the question is asking for the probability of a club <u>or</u> a jack, we know that we have to add the two probabilities. The first probability is that of picking a club, which is 13/52. The probability of picking a jack (be sure not to overlap; don't double count the jack of clubs) is 3/52. Adding these two gives us 13/52+3/52=16/52, which simplifies to 4/13.

3 0

3 years ago

A tree is growing at a rate of 36 feet every 3

Airida [17]

well, we know the tree is growing at a rate of 36 feet every 3 years, how many feet is that in 1 year alone? well, simply 36/3 = 12.

so the tree is growing 12 feet per year.

when it was first planted, it was 5 feet tall, and then after every year, we have to add 12 feet subsequently, let's do that

year 1 .......... 5 + 12(1)

year 2 .......... 5 + 12(2)

year 3 .......... 5 + 12(3)

year 4 .......... 5 + 12(4)

year 5 .......... 5 + 12(5)

year 6 .......... 5 + 12(6)

year 7 .......... 5 + 12(7)

year y .......... 5 + 12(y)

$\bf \stackrel{\textit{height}}{h}~~=~~\stackrel{\textit{initial height}}{5}~~+~~\stackrel{\textit{yearly rate}}{12y} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{tree is now}}{77} = 5 + 12y\implies 72=12y\implies \cfrac{72}{12}=y\implies \boxed{6=y}$

3 0

3 years ago