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3 years ago

Use the graphs below to find the slope of each line (0,8),(8,1)

Mathematics

1 answer:

nignag [31]3 years ago

6 0

Answer:

slope= -7/8

Step-by-step explanation:

y2-y1=1-8=-7

x2-x1=8-0=8

-7/8

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-8 times 0.09 times -0.5

Black_prince [1.1K]

The answer is 0.36. I don't think u need to explain why, right ;)

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4 years ago

Read 2 more answers

Give me three sides to a right triangle. Explain how you can use the Pythagorean theorem to know that your three sides will crea

dsp73

The three numbers: 12, 35, 37

Applying these number to Pythagorean theorem…
a2+b2=c2
So,
12^2+35^2=37^2
144+1225=1369
1369=1369

As you can see both are equal (1369=1369), so we have a right angle triangle. If both sides are not equal then it is not classified as a right angle triangle.

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2 years ago

Make a table of ordered pairs for the equation.

stira [4]

$y \: = \: \frac{ - 1}{3} x \: + \: 1$
When y = 0,
$0 \: = \: \frac{ - 1}{3} x \: + \: 1 \\ \frac{ 1}{3} x \: = \: 1 \\ x \: = \: 3$
When x = 0,
$y \: = \: 0 \: + \: 1 \\ y \: = \: 1$
Therefore, Ordered pairs are (0,1), (3,0), etc.

4 0

3 years ago

67% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 48 owned dogs are r

oksian1 [2.3K]

Answer:

a) Probability that exactly 29 of them are spayed or neutered = 0.074

b) Probability that at most 33 of them are spayed or neutered = 0.66

c) Probability that at least 30 of them are spayed or neutered = 0.79

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered = 0.574

Step-by-step explanation:

This is a binomial distribution question

probability of having a spayed or neutered dog, p = 0.67

probability of having a dog that is not spayed or neutered, q = 1 - 0.67

q = 0.23

sample size, n = 48

According to binomial distribution formula:

$P(X=r) = nCr p^r q^{n-r}$

where $nCr = \frac{n!}{(n-r)! r!}$

a) Probability that exactly 29 of them are spayed or neutered

$P(X= 29) = 48C29 * 0.67^{29} * 0.23^{19}\\P(X=29) = 0.074$

b) Probability that at most 33 of them are spayed or neutered

$P(X \leq 33) =1 - P(X > 33)\\P(X \leq 33) =1 - 0.34\\P(X \leq 33) = 0.66$

c) Probability that at least 30 of them are spayed or neutered

$P(X \geq 30) = 1 - P(x < 30)\\P(X \geq 30) = 1 - 0.21\\P(X \geq 30) = 0.79$

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered.

$P(28 \leq X \leq 33) = P(X=28) + P(X=29) + P(X=30) + P(X=31) + P(X=32) + P(X=33)\\P(28 \leq X \leq 33) = 0.053 + 0.074 + 0.095 + 0.112 + 0.121 + 0.119\\P(28 \leq X \leq 33) = 0.574$

8 0

3 years ago

Yo, which 288in^2 should I choose? I have a feeling AMS is messing with me at this point.

Oduvanchick [21]

That is weird i think both answers would be the same they probably just put the same answer choice accidentally.

Hope this was helpful :-) <span />

6 0

3 years ago