HURRRRRRYYYYYYY 1 QUESTION EXPERTS/ACE HELP QUICK!!!!!

Bess and Gina earn $45 per week for delivering flowers. Bess worked for x weeks and earned an additional total bonus of $20. Gin

Inessa05 [86]

Answer:

First option: $45x + 20 + 45y$

Step-by-step explanation:

The missing options are:

$45x + 20 + 45y\\65x + 45y\\45x - 20 + 45y\\\\45x + 65y$

We know that "x" represents the number of weeks Bess worked and "y" the number of weeks Gina worked.

According to the data given in the exercise, Gina and Bess earn $45 per week for delivering flowers, this is:

$45x+45y$

Since earned an additional total bonus of $20, then the total money in in dollars that Bess and Gina earned for delivering flowers can be shown with the following expression:

$(45x+20)+45y=45x+20+45y$

Notice that this expression matches with the one shown in the first option.

8 0

3 years ago

Mary,Tom,Sophia & Isaac share minutes on their cell phone plan. Mary used 1/4 of the minutes,Tom used 0.5 of the minutes,and

nataly862011 [7]

Hello there! So, 1/4 is equivalent to 0.25, which is 25% in percent form. 0.5 is 50% and 0.2 is 20% in percent form. 50 + 20 + 25 is 95. 100 - 95 is 5. Issac can use 5% of the minutes on the cell phone plan. The answer is D: 5%.

5 0

3 years ago

the volume of a rectangular prism can be computed using the formula V=lwh what is the width of a prism that has a volume of 240

Anuta_ua [19.1K]

The width would be 4 centimeters

6 0

3 years ago

5

nika2105 [10]

Answer:

idk

Step-by-step explanation:

8 0

3 years ago

Suppose it is known that the distribution of purchase amounts by customers entering a popular retail store is approximately norm

iragen [17]

Answer:

a. 0.691

b. 0.382

c. 0.933

d. $88.490

e. $58.168

f. 5th percentile: $42.103

95th percentile: $107.897

Step-by-step explanation:

We have, for the purchase amounts by customers, a normal distribution with mean $75 and standard deviation of $20.

a. This can be calculated using the z-score:

$z=\dfrac{X-\mu}{\sigma}=\dfrac{85-75}{20}=\dfrac{10}{20}=0.5\\\\\\P(X$

The probability that a randomly selected customer spends less than $85 at this store is 0.691.

b. We have to calculate the z-scores for both values:

$z_1=\dfrac{X_1-\mu}{\sigma}=\dfrac{65-75}{20}=\dfrac{-10}{20}=-0.5\\\\\\z_2=\dfrac{X_2-\mu}{\sigma}=\dfrac{85-75}{20}=\dfrac{10}{20}=0.5\\\\\\\\P(65$

The probability that a randomly selected customer spends between $65 and $85 at this store is 0.382.

c. We recalculate the z-score for X=45.

$z=\dfrac{X-\mu}{\sigma}=\dfrac{45-75}{20}=\dfrac{-30}{20}=-1.5\\\\\\P(X>45)=P(z>-1.5)=0.933$

The probability that a randomly selected customer spends more than $45 at this store is 0.933.

d. In this case, first we have to calculate the z-score that satisfies P(z<z*)=0.75, and then calculate the X* that corresponds to that z-score z*.

Looking in a standard normal distribution table, we have that:

$P(z$

Then, we can calculate X as:

$X^*=\mu+z^*\cdot\sigma=75+0.67449\cdot 20=75+13.4898=88.490$

75% of the customers will not spend more than $88.49.

e. In this case, first we have to calculate the z-score that satisfies P(z>z*)=0.8, and then calculate the X* that corresponds to that z-score z*.

Looking in a standard normal distribution table, we have that:

$P(z>-0.84162)=0.80$

Then, we can calculate X as:

$X^*=\mu+z^*\cdot\sigma=75+(-0.84162)\cdot 20=75-16.8324=58.168$

80% of the customers will spend more than $58.17.

f. We have to calculate the two points that are equidistant from the mean such that 90% of all customer purchases are between these values.

In terms of the z-score, we can express this as:

$P(|z|$

The value for z* is ±1.64485.

We can now calculate the values for X as:

$X_1=\mu+z_1\cdot\sigma=75+(-1.64485)\cdot 20=75-32.897=42.103\\\\\\X_2=\mu+z_2\cdot\sigma=75+1.64485\cdot 20=75+32.897=107.897$

5th percentile: $42.103

95th percentile: $107.897

5 0

3 years ago