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ArbitrLikvidat [17]
3 years ago
15

Which is it, true or false?

Mathematics
1 answer:
Ugo [173]3 years ago
4 0

Answer: I think it's true

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I dont understand help me
DochEvi [55]
1.5x=3

First you want to get x by itself so divide both sides by 1.5

3/1.5=2
x=2

4 0
4 years ago
. The diameter of Bubba's bubble is
Oksana_A [137]

Answer:

10.22

Step-by-step explanation:

2.45 * 10 =20.45 - 3.67 * 10 =30.67

30.67 - 20.45

30 - 20 = 10

.67 - .45 = .22

30.67 - 20.45 = 10.22

8 0
3 years ago
you deposit $7,900 in a money-market account that pays an annual interest rate of 4.3%. the interest is compounded quarterly. ho
Sladkaya [172]
<span><span>anonymous </span> 3 years ago</span>we are solving for A, A is the future amount A=? P is the present value, so we were told 7,900, so p=7900 r is the interest rate, so 4.3%, but in decimal form so r=0.043 k is the how often it is 'compounded', it says quarterly, so k=4 n is how long, 3 years, so n=3 <span>A=P(1+<span>rk</span><span>)<span>n∗k</span></span></span> plug in the values
<span>A=8981.51</span>
4 0
3 years ago
The side of a square is x, and the area is A. Express A in terms of x. Draw the graph which shows how A depends on x.
Alexandra [31]

a=x squared or x times itself

4 0
3 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
4 years ago
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