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frez [133]
3 years ago
14

Can someone help me?

Mathematics
1 answer:
a_sh-v [17]3 years ago
8 0

Answer:

f^-1 (x) = - 6x-1 / 10 + 9x

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A large box of jelly beans weighs 96.58 lb the jelly beans are evenly divided into 22 bags what is the weight in pounds of the j
Vadim26 [7]

Answer:

The weight in each bag is 4.39lb.

Step-by-step explanation:

To solve this, you simply need to divide 96.58 by 22. The reason you do this is so that way you can divide all of the weight into 22 bags.

96.58 ÷ 22 = 4.39

This means that the weight in each bag is 4.39lb.

4 0
2 years ago
mrs lim had some money she spent 4/7 of it on necklace and 1/6 of her remainig money on a bracelet she then had $990 left how mu
zmey [24]

Answer:

she spent $1782

Step-by-step explanation:

first we have to find out her total money then minus 990

to get that 990 /5 then multipliedby 6 is 1188

1188 over 3 then times by 7 is 2772

2772 is the total amount of money she had. minus what she had left which is 990 she has spent 1782 dollars

4 0
2 years ago
How would you read this inequality? Answer in a sentence. x<21
Lubov Fominskaja [6]

Answer:

x is less than twenty one

5 0
2 years ago
Does anyone have a discord server for homework help?​
Deffense [45]

Answer:

I do....................................

7 0
2 years ago
There are 20 machines in a factory. 7 of the machines are defective.
adelina 88 [10]

Answer:

0.1225

Step-by-step explanation:

Given

Number of Machines = 20

Defective Machines = 7

Required

Probability that two selected (with replacement) are defective.

The first step is to define an event that a machine will be defective.

Let M represent the selected machine sis defective.

P(M) = 7/20

Provided that the two selected machines are replaced;

The probability is calculated as thus

P(Both) = P(First Defect) * P(Second Defect)

From tge question, we understand that each selection is replaced before another selection is made.

This means that the probability of first selection and the probability of second selection are independent.

And as such;

P(First Defect) = P (Second Defect) = P(M) = 7/20

So;

P(Both) = P(First Defect) * P(Second Defect)

PBoth) = 7/20 * 7/20

P(Both) = 49/400

P(Both) = 0.1225

Hence, the probability that both choices will be defective machines is 0.1225

4 0
3 years ago
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