Yes, it makes sense to represent the relationship between the amount saved and the number of months with one constant rate. The relationship is 35x+ 100.

<h3>Relationship between the amount saved and the number of months</h3>

After 1 month, Jane will saved=$35+$100

After 1 month, Jane has saved=$135

Hence,

Let x represent the number of months

Since every month she saved $35 which inturn means that in x number of months she can save 35x. Based on this the relationship between the amount saved and the number of months is 35x +100.

Therefore it makes sense to represent the relationship between the amount saved and the number of months with one constant rate. The relationship is 35x+ 100.

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4 0

1 year ago

PLEASE HELP!! Find the surface area of each figure to the nearest tenth. Show your work.

hram777 [196]

Surface area = 2(ab+bc+ac)
a=5.2 ft, b=2.4 ft, c=3.5ft

Surface area = 2(5.2 * 2.4 + 2.4 * 3.5 + 3.5 * 5.2) =
2(12.48 + 8.4 + 18.2) =
2 * 39.08 =
78.16 ≈ 78.2 ft² ← <span>to the nearest tenth</span>

7 0

3 years ago

A supplier delivers an order for 20 electric toothbrushes to a store. By accident, three of the electric toothbrushes are defect

Vladimir79 [104]

The probability that the first two electric toothbrushes sold are defective is 0.016.

The probability of an event, say E occurring is:

$P(E)=\frac{n(E)}{N}$

Here,

n (E) = favorable outcomes

N = total number of outcomes

Let X = the number of defective electric toothbrushes sold.

The number of electric toothbrushes that were delivered to a store is n = 20.

The number of defective electric toothbrushes is x = 3.

The number of ways to select two toothbrushes to sell from the 20 toothbrushes is:

$(\left {{20} \atop {2}} \right. )$ $=\frac{20!}{2!(20-2)!} =\frac{20!}{2!18!} =\frac{20*19*18!}{2!*18!} = 190$

The number of ways to select two defective toothbrushes to sell from the 3 defective toothbrushes is:

$(\left {{3} \atop {2}} \right. )$ $=\frac{3!}{2!(3-2)!} =\frac{3!}{2!1!} =\frac{3*2!}{1!2!} =3$

Compute the probability that the first two electric toothbrushes sold are defective as follows:

P (Selling 2 defective toothbrushes) = Favorable outcomes ÷ Total no. of outcomes

$\frac{3}{190}\\ =0.01579\\=0.016$

Thus, the probability that the first two electric toothbrushes sold are defective is 0.016.

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4 0

1 year ago