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HELP HELP HELP Quick algebra 1 questions

beks73 [17]

Hello:

Okay, this problem should be taken in a series of steps:

What is the slope's formula?

$\rm\hookrightarrow \frac{y_2-y_1}{x_2-x_1}$

(x₁,y₁) -- first point
(x₂,y₂) -- second point

Okay, now let's use this knowledge

10. (5,-19) and (-5,21)

$Slope = \frac{21--19}{-5-5} =\frac{40}{-10}=-4$

11. (12,1) and (12, -1)

$Slope = \frac{-1-1}{12-12} =\frac{-2}{0}$

12. (8,7) and (4,7)

$Slope = \frac{7-7}{4-7} =0$

Answer:

Question 10: -4
Question 11: undefined
Question 12: 0

Hopefully that helps!

5 0

2 years ago

The perimeter of the scalene triangle is 54.6 cm. Which equation can be used to find the value of b if side a measures 8.7 cm?

77julia77 [94]

A scalene triangle is a triangle that has 3 sides that are all different lengths.

Let these letters represent the problem:

a = 8.7

b = side 2

c = side 3

P = 54.6

To find the perimeter, we just need to add all the sides [ P = a + b + c ]

So, put what we have in the formula above.

54.6 = 8.7 + b + c

Best of Luck!

7 0

3 years ago

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In 1997, the American Diabetes Association (ADA) and the federal government lowered the standard for diagnosing diabetes from a

Klio2033 [76]

Answer:

10 percent

Step-by-step explanation:

- Before 1997, people with a minimum FBGL of 140mg/dl were diagnosed as diabetic

- After 1997, people with a minimum FBGL of 126mg/dl were diagnosed as diabetic

- What is the proportion of people who were not considered diabetic before 1997 but are now considered diabetic (after 1997)?

140 - 126 = 14

In percentage, this proportion is = 14/140 × 100

= 10%

Hence, 10% of the people who were not considered diabetic before 1997 are now or will now be considered diabetic.

3 0

3 years ago

Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we

Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N( $\mu = 7 lbs , \sigma^{2} = 0.875^{2} lbs$ )

The standard normal z distribution is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, X bar = sample mean weight

n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{7.5-7}{\frac{0.875}{\sqrt{4} } }$ ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

8 0

3 years ago

You make $8.95 an hour working at a coffee shop. If you earn $107.40 how many hours did you work?

puteri [66]

12 hours (107.40 / 8.95)

7 0

3 years ago

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