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lina2011 [118]
1 year ago
15

in 448 244 how is the relationship between the first pair of 4s the same as the relationship between the second pair of 4s

Mathematics
2 answers:
sineoko [7]1 year ago
8 0
Well there similar because your adding 10 more or 10× since from ones to the tens is 10×
Aliun [14]1 year ago
6 0

Answer:

In both pairs the 4 on the left is ten times the amount of the 4 on the right.

Step-by-step explanation:

The number is : 448 244

The relationship between the first pair of 4's is the same as the relationship between the second pair of 4's.

In both pairs the 4 on the left is ten times the amount of the 4 on the right.

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G = x-c/x solve for x
Elan Coil [88]

Answer:

x=\frac{c}{1-g}

Step-by-step explanation:

Step 1: Multiply both sides by x,

 gx=−c+x

Step 2: Add -x to both sides.

gx+−x=−c+x+−x

gx−x=−c

Step 3: Factor out variable x.

x(g−1)=−c

Step 4: Divide both sides by g-1.

\frac{x(g-1)}{g-1} =\frac{-c}{g-1\\}

x=\frac{c}{1-g}

Hope this helps!

5 0
2 years ago
Read 2 more answers
The NEC permits 20% of the cross sectional area in a wireway to be occupied by conductors.for an 8"×8"×6' long wireway the condu
Zanzabum

Answer:

Step-by-step explanation:

Sjaoa

8 0
1 year ago
at an amusement park, Robin bought a t-shirt for $8 and 5 ticket for ride. she spent a total of $23. How much did each ticket co
Elanso [62]

Each ticket cost $3

8+5x=23

5x=15

x=3

3 0
2 years ago
Read 2 more answers
According to a study in a medical journal, 202 of a sample of 5,990 middle-aged men had developed diabetes. It also found that m
tekilochka [14]

Answer:

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Has diabetes.

Event B: Is very active.

Probability of having diabetes:

To find this probability, we take in consideration that:

It also found that men who were very active (burning about 3,500 calories daily) were a fourth as likely to develop diabetes compared with men who were sedentary. Assume that one-fifth of all middle-aged men are very active, and the rest are classified as sedentary.

So the probability of developing diabetes is:

x of 4/5 = x of 0.8(not active)

x/4 = 0.25x of 1/5 = 0.2(very active). So

P(A) = 0.8x + 0.25*0.2x = 0.85x

Probability of developing diabetes while being very active:

0.25x of 0.2. So

P(A \cap B) = 0.25x*0.2 = 0.05x

What is the probability that a middle-aged man with diabetes is very active?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.05x}{0.85x} = \frac{0.05}{0.85} = 0.0588

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

4 0
1 year ago
A rectangular field is enclosed by 360 feet of fencing. What is the length, in feet, of the field if it’s length is 6 feet more
Orlov [11]

Answer:

The Field's length is 93ft

Step-by-step explanation:

P=2L+2w

360=2(w+6)+2w

360=2w+12+2w

360=4w+12

348=4w

348/4=w

87=w

L=w+6

L=87+6

L=93

7 0
2 years ago
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