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Mashcka [7]
3 years ago
7

Please help!! I have 5 mins

Mathematics
2 answers:
gtnhenbr [62]3 years ago
7 0

Answer:

3/2

Step-by-step explanation:

Just count the rise over run

kotykmax [81]3 years ago
3 0

Answer:

\frac{3}{2}

Step-by-step explanation:

Use rise/run.

Between points, it rises by 3, then runs laterally by 2.

Thus, 3/2.

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Can someone help me please ???????
Inessa [10]

Answer:

d=1

Step-by-step explanation:

Factor d^2 - 2d - 8 into (d-4)(d+2)

Move -2/d+2 onto the other side, changing it into 2/d+2.

\frac{-3d}{(d-4)(d+2)} +\frac{3}{d-4}+\frac{2}{d+2} = 0

Let the empty side equal zero.

Add 3/d-4 and 2/d+2

\frac{5d-2}{(d+2)(d-4)}

Then add that to -3d.

\frac{2d-2}{(d+2)(d+4)}

Multiply 0 by d+2 and d-4 to get 2d-2 by itself.

2d=2

d=1

I can't edit the equation, but it's d-4 althroughout the question. Sorry for being so slow.

6 0
3 years ago
Write a linear equation for the information given below.
suter [353]

Answer:

below

Step-by-step explanation:

that is the procedure above

8 0
3 years ago
The resting heart rate for an adult horse should average about µ = 47 beats per minute with a (95% of data) range from 19 to 75
KatRina [158]

Answer:

a. 0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. 0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. 0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

Step-by-step explanation:

Empirical Rule:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean:

\mu = 47

(95% of data) range from 19 to 75 beats per minute.

This means that between 19 and 75, by the Empirical Rule, there are 4 standard deviations. So

4\sigma = 75 - 19

4\sigma = 56

\sigma = \frac{56}{4} = 14

a. What is the probability that the heart rate is less than 25 beats per minute?

This is the p-value of Z when X = 25. So

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

Z = \frac{25 - 47}{14}

Z = -1.57

Z = -1.57 has a p-value of 0.0582.

0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. What is the probability that the heart rate is greater than 60 beats per minute?

This is 1 subtracted by the p-value of Z when X = 60. So

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

Z = \frac{60 - 47}{14}

Z = 0.93

Z = 0.93 has a p-value of 0.8238.

1 - 0.8238 = 0.1762

0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. What is the probability that the heart rate is between 25 and 60 beats per minute?

This is the p-value of Z when X = 60 subtracted by the p-value of Z when X = 25. From the previous two items, we have these two p-values. So

0.8238 - 0.0582 = 0.7656

0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

3 0
3 years ago
There are 15 candidates for 4 job positions.
Veronika [31]

Answer:

32760 ways

Step-by-step explanation:

Given

Number of Candidates = 15

Job Positions = 4

Required:

Number of outcomes

This question represent selection; i.e. selecting candidates for job positions;

This question can be solved in any of the following two ways

Method 1.

The first candidate can be chosen from any of the 15 candidates

The second candidate can be chosen from any of the remaining 14 candidates

The third candidate can be chosen from any of the remaining 13 candidates

The fourth candidate can be chosen from any of the remaining 12 candidates

Total Possible Selection = 15 * 14 * 13 * 12

<em>Total Possible Selection = 32760 ways</em>

<em></em>

Method 2:

This can be solved using permutation method which goes thus;

nPr = \frac{n!}{(n-r)!}

Where n = 15 and r = 4

So;

nPr = \frac{n!}{(n-r)!} becomes

15P4 = \frac{15!}{(15-4)!}

15P4 = \frac{15!}{11!}

15P4 = \frac{15*14*13*12*11!}{11!}

15P4 = 15*14*13*12

15C4 = 32760

<em>Hence, there are 32760 ways</em>

8 0
4 years ago
Find one counter example to show that the conjecture is false. angle 1 and angle 2 are​ supplementary, so one of the angles is a
Mrac [35]

Answer:

B. m ∠ 1 = 90° and m ∠ 2 = 90°

Step-by-step explanation:

For most situations, the conjecture would probably be true, but there is one exception that makes this statement false.

When two right angles are supplementary, none of them is acute.

For an angle to be acute it needs to be lesser than 90°, and for a pair of angles to be supplementary they should add up to exactly 180°.

With a pair of right angles (90° each), their sum adds up to 180° but neither of them are acute.

Therefore, the answer is B. m ∠ 1 = 90° and m ∠ 2 = 90°

4 0
4 years ago
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