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2 years ago

7

Denise is making bean bags. She has 6.4 pounds of beans.

1 answer:

dezoksy [38]2 years ago

5 0

Answer:

a) she be able to make 8 bags

b) she can make 16 bags.

Step-by-step explanation:

According to the exercise info Denise has 6.4 pounds of beans

<u>Solving question a </u>

⇒ if she makes each bean bag of 0.8 pounds and she has 6.4 pounds of beans, we can express this by using the three simply rule she needs to puts 6.4 pounds of beans in x quantity of bags of 0.8 pouns, so doing the division: 6. 4 pounds of beans / 0.8 pounds of beans per bag= 8 bags of 0.8 pounds of beans.

B- If she decides to make mini bean bags that are half as heave as the ohers bags, so now Denise is going to make bags of 0.8/2 = 0.4 pounds.

Doing the same procedure as question a, we can say that she has 6.4 pounds of beans and she needed to put these amount in a x quantity of 04 pounds bags. Doing the following division ⇒ 6.4 pounds of beans/ 0.4 pounds of beans per bag= 16 bags

In conclusion she is going to make 16 bags of 0.4 pounds of beans.

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olga2289 [7]

Answer:

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Step-by-step explanation:

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1 year ago

PLEASE HELP! A horse-drawn carriage tour company has found that the number of people that take their tour depends on the price c

Natasha_Volkova [10]

A is the correct answers

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1 year ago

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Mandarinka [93]

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1 year ago

t Technodynamics, Inc., a randomly-selected hiring committee of 3 people is formed from a group of 4 employees in marketing an

atroni [7]

Answer: a) 0.25, b) 0.78, c) 0.71

Step-by-step explanation:

Since we have given that

Number of employees in marketing = 4

Number of employees in management = 7

We need to hire committee of 3 people.

a) Find the probability that the committee has exactly 2 employees from marketing.

So, Probability becomes

$\dfrac{^4C_2\times ^7C_1}{^{11}C_3}\\\\=\dfrac{42}{165}\\\\=0.25$

b) Find the probability that the committee has at least one employee from marketing.

$P(x\geq 1)=1-P(x=0)\\\\P(x\geq 1)=1-\dfrac{^7C_3}{^{11}C_3}=1-0.212=0.78$

c) Find the probability that the committee has at most one employee from management.

$P(x\leq 1)=P(x=0)+P(x=1)\\\\P(x\leq 1)=\dfrac{^7C_3}{^{11}C_3}+\dfrac{^7C_2\times ^4C_1}{^{11}C_3}\\\\P(x\leq 1)=0.21+0.50=0.71$

Hence, a) 0.25, b) 0.78, c) 0.71

8 0

1 year ago