What is the area of the reduced rug, in square feet?

I need help with this pleaseee

Elodia [21]

Answer:

34, D

Step-by-step explanation:

Change 4 and 1/4 to an improper fraction

4 1/4 = 17/4

Now, you would divide 17/4 by 1/8...

17/4 divided by 1/8

Reciprocate the fraction 1/8 and change the division sign to multiplication to get...

17/4 multiplied by 8/1

= 136 / 4

Simplify

136 / 4

= 34

So, you can grow 34 plants.

Hope this helps!

7 0

3 years ago

A local university wanted to understand what students prefer to eat during finals. They asked 1000 students, "Do you prefer chic

olga55 [171]

Answer:

$P(\text{Pizza}\ |\ \text{Female})=0.225$

Step-by-step explanation:

The two-way frequency table is attached below.

We have to calculate the probability of, a person chosen at random prefers pizza given that they are female, i.e $P(\text{Pizza}\ |\ \text{female})$

This is a conditional probability.

We know that,

$P(A\ |\ B)=\dfrac{P(A\ \cap\ B)}{P(B)}$

So,

$P(\text{Pizza}\ |\ \text{Female})=\dfrac{P(\text{Pizza}\ \cap\ \text{Female)}}{P({\text{Female}})}$

From the table,

$P(\text{Pizza}\ \cap\ \text{Female)}=\dfrac{119}{1000}$

$P({\text{Female}})=\dfrac{530}{1000}$

Putting the values,

$P(\text{Pizza}\ |\ \text{Female})=\dfrac{P(\text{Pizza}\ \cap\ \text{Female)}}{P({\text{Female}})}=\dfrac{\frac{119}{1000}}{\frac{530}{1000}}=0.225$

8 0

3 years ago

It has a time to failure distribution which is normal with a mean of 35,000 vehicle miles and a standard deviation of 7,000 vehi

Triss [41]

Answer:

The designed life should be of 21,840 vehicle miles.

Step-by-step explanation:

Normal Probability Distribution

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

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 35,000 vehicle miles and a standard deviation of 7,000 vehicle miles.

This means that $\mu = 35000, \sigma = 7000$

Find its designed life if a .97 reliability is desired.

The designed life should be the 100 - 97 = 3rd percentile(we want only 3% of the vehicles to fail within this time), which is X when Z has a p-value of 0.03, so X when Z = -1.88.

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

$-1.88 = \frac{X - 35000}{7000}$