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salantis [7]
3 years ago
11

Joe Traffic gained 986 yards during football season.Ziggy Fumble lost 118 yards during the season.What was the difference in the

ir yardage gains
Mathematics
1 answer:
Bond [772]3 years ago
6 0

Joe is positive 986 yards

Ziggy is negative 118 yards

the total difference is 986 +118 = 1,104 yards

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Please help this is i-ready!
KengaRu [80]

Answer: It will take 3 days.

Step-by-step explanation:

1 km = 1000m

12 km = 12,000m

10 * 400m  = 4000m

12,000m / 1000m = 3

So, It will take 3 Days.

I hope I helped you with this I-Ready. Have a Great day!

3 0
3 years ago
Read 2 more answers
A number is chosen at random from the first 100 positive integers. find the probability that the number is divisible by 5. 1/10
Arlecino [84]

The probability that the number chosen is divisible by 5 is; 1/5.

<h3>What is the probability that the number chosen is divisible by 5?</h3>

It follows from the task content that the number of possible outcomes in the probability event is; 100. Additionally, the number of required outcomes is 20 as only 20 numbers are divisible by 5.

Ultimately, the required probability is; 20/100 = 1/5.

Read more on probability;

brainly.com/question/7965468

$SPJ1

8 0
2 years ago
The sum of 3x^2+4x-2 and x^2-5x+3 is
Diano4ka-milaya [45]

The sum would be

4x^2-x+1

add like terms.

3x^2+x^2

4x^2

4x+-5x

-x

3-2

1


3 0
3 years ago
Read 2 more answers
A new sample of 225 employed adults is chosen. Find the probability that less than 7.1% of the individuals in this sample hold m
arsen [322]

Answer:

The probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

Step-by-step explanation:

Let <em>X</em> = number of individuals in the United States who held multiple jobs.

The probability that an individual holds multiple jobs is, <em>p</em> = 0.13.

The sample of employed individuals selected is of size, <em>n</em> = 225.

An individual holding multiple jobs is independent of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 225 and <em>p</em> = 0.13.

But since the sample size is too large Normal approximation to Binomial can be used to define the distribution of proportion <em>p</em>.

Conditions of Normal approximation to Binomial are:

  • np ≥ 10
  • n (1 - p) ≥ 10

Check the conditions as follows:

np=225\times 0.13=29.25>10\\n(1-p)=225\times (1-0.13)=195.75>10

The distribution of the proportion of individuals who hold multiple jobs is,

p\sim N(p, \frac{p(1-p)}{n})

Compute the probability that less than 7.1% of the individuals in this sample hold multiple jobs as follows:

P(p

*Use a <em>z</em>-table.

Thus, the probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

7 0
3 years ago
Which expression represents five times the quotient of two numbers
kotykmax [81]

5 times the quotient of 2 numbers is 5(r/t)

Answer is the third option

5(r/t)

7 0
3 years ago
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