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Elanso [62]
2 years ago
12

You need to make gift bags for a party with the same number of lollipops and chocolate in each bag. One package of lollipops has

24 pieces. One bag of chocolate has 20 bars. You need to use all the lollipops and all the chocolate bars. Part 2 what is the greatest number of gift bags that you can make containing and equal number of items?
Mathematics
2 answers:
Ray Of Light [21]2 years ago
8 0

120 :)))))))))))))))))))))))))))

sveticcg [70]2 years ago
3 0
The greatest number of gift bags you can create is 120 as the lowest common multiple of both numbers is 120
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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

5th percentile: $42.103

95th percentile: $107.897

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A construction company can remove
kolbaska11 [484]

Answer:

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NEED ANSWER ASAP!
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Answer:

t=0.57h

Step-by-step explanation:

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