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torisob [31]
4 years ago
9

Noor brought 21 sheets of stickers. She gave 1/3 of a sheet to each of the 45 students at recess. She wants to give teachers 1 1

/2 sheets each. What inequality can describe the number of teachers ,t, Noor could give sheets of stickers too?
Mathematics
1 answer:
Tamiku [17]4 years ago
6 0

Answer:

\frac{3}{2}t+15\leq 21

Step-by-step explanation:

Let t represent the number of teachers.

We have been given that Noor gave 1/3 of a sheet to each of the 45 students at recess.

Let us find the number of sheets given to students by Noor by multiplying the total number of students by the part of sheet given to each student.

\text{Number of sheets given to 45 students}=45\times \frac{1}{3}

\text{Number of sheets given to 45 students}=15

So, Noor gave 15 sheets to 45 students.

We have been given that Noor wants to give 1\frac{1}{2}=\frac{3}{2} sheet to each teachers, so the number of sheets given to t teachers will be \frac{3}{2}t.

As Noor has bought 21 sheets, so the number of sheets given to students and t teachers will be less than or equal to 21.

We can represent this information in an inequality as:

\frac{3}{2}t+15\leq 21

Therefore, the inequality \frac{3}{2}t+15\leq 21 represents the number of teachers ,t, Noor could give sheets of stickers too.

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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

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95th percentile: $107.897

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