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FromTheMoon [43]
3 years ago
11

What is 65 less then j in a algebraic expression

Mathematics
2 answers:
const2013 [10]3 years ago
8 0
65-j would be the answer
hope i helped!
neonofarm [45]3 years ago
4 0
65 less than J in an algebraic expression would be " 65-j "
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What is the type slope?<br> Please help
nadya68 [22]

Answer:

its positive

Step-by-step explanation:

because the line going upward

6 0
2 years ago
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Which statements are true regarding undefinable terms in geometry? Check all that apply.
Vanyuwa [196]
I do not have your choices but all the statements that I can think of is the following: A point in the x,y form has two dimensions, a plane has a definite beginning and end, the line has one dimension or length, and finally a point consists of an infinite set of the lines  <span />
7 0
3 years ago
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A) how many ways are there to choose 10 players to take the field?
Kruka [31]

A. C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.

B. P(13,10)= 13! =13! =13·12·11·10·9·8·7·6·5·4. (13−10)! 3!

C. f there is exactly one woman chosen, this is possible in C(10, 9)C(3, 1) =

10! 3!

9!1! 1!2! 10! 3!

8!2! 2!1! 10! 3!

7!3! 3!0!

= 10 · 3 = 30 ways; two women chosen — in C(10,8)C(3,2) =

= 45·3 = 135 ways; three women chosen — in C(10, 7)C(3, 3) =

= 10·9·8 ·1 = 120 ways. Altogether there are 30+135+120 = 285 1·2·3

<span>possible choices.</span><span> 
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5 0
3 years ago
Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean
Mamont248 [21]

Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 50, \sigma = 6

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

Z = \frac{X - \mu}{\sigma}

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

Z = \frac{X - \mu}{\sigma}

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

Z = \frac{X - \mu}{\sigma}

Z = \frac{67 - 50}{6}

Z = 2.83

Z = 2.83 has a pvalue 0.9977

X = 60

Z = \frac{X - \mu}{\sigma}

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

Z = \frac{X - \mu}{\sigma}

Z = \frac{76 - 50}{6}

Z = 4.33

Z = 4.33 has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

Z = \frac{X - \mu}{\sigma}

-1.28 = \frac{X - 50}{6}

X - 50 = -1.28*6

X = 42.32

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

Z = \frac{X - \mu}{\sigma}

-1.28 = \frac{X - 50}{6}

X - 50 = -1.28*6

X = 42.32

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

Z = \frac{X - \mu}{\sigma}

1.28 = \frac{X - 50}{6}

X - 50 = 1.28*6

X = 57.68

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

5 0
3 years ago
A customer sent in a $230 check to pay off an overdue bill but failed to pay the interest that accrued over the 45 days after th
viva [34]
$230+($45x$0.42=248.90

That is the equation to solve it.
4 0
3 years ago
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