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3 years ago

What sum is represented by the arrow diagram?

Mathematics

1 answer:

Julli [10]3 years ago

4 0

Answer:

Would it be 1?

Step-by-step explanation:

I'm not that sure but it seems it goes up to 4 so 4- -3 = 1. Exuse me if my wrong..

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What are the solutions to the equation (m + 1)2 +1 = 5? (1 point)

ale4655 [162]

Answer:

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3 years ago

A beverage is made by mixing 2 parts of water with 3 parts of fruit juice. How many parts of water are mixed with 1 part of frui

MrRissso [65]

I'm not sure... ratio is 2:3, so it should be 2/3

8 0

3 years ago

Read 2 more answers

HEY YALL CAN U HELP PLZ THANK U

maria [59]

Answer: D. 2

Step-by-step explanation:

Given

y-1=2(x--2)

Change signs

y-1=2(x+2)

Expand parenthesis

y-1=2x+4

Add 1 on both sides to isolate "y"

y-1+1=2x+4+1

y=2x+4

So the slope is 2

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7 0

3 years ago

Function g(x) has a removable discontinuity at x = c, where g(c) is undefined. Which statement describes the one-sided limits at

Norma-Jean [14]

Answer:

Step-by-step explanation:

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

krek1111 [17]

Answer:

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

Step-by-step explanation:

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 = number of defective electric toothbrushes sold.

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

Number of defective electric toothbrushes is, x = 3.

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

${20\choose 2}=\frac{20!}{2!(20-2)!}=\frac{20!}{2!\times 18!}=\frac{20\times 19\times 18!}{2!\times 18!}=190$

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

${3\choose 2}=\frac{3!}{2!(3-2)!}=\frac{3!}{2!\times 1!}=3$

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

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

$=0.01579\\\approx0.016$

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

8 0

3 years ago