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Maslowich
3 years ago
6

I want solution 1 and 2 with the steps to know the final answer

Mathematics
1 answer:
Delicious77 [7]3 years ago
4 0
Question 1:

Slope = 1/5

y = mx + c 
y = 1/5 x + c
at point (5, -1)
-1 = 1/5 (5) + c
- 1= 1 + c
c = - 2

y = 1/5x - 2
5y = x - 10

Question 2:
slope = (9-5)/(3-1)
Slope = 2

y = mx + c
y = 2x + c

at point (1, 5)
5 = 2(1) + c
c = 5 - 2
c = 3

y = 2x + 3


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Can the distributive property be used to rewrite 6(9 - 4) ?
alukav5142 [94]

Answer:

yes distribute . . .

Step-by-step explanation:

the 6 to the 9 and the 4

6(9) - 6(4)=30

5 0
3 years ago
Kim purchased a desktop computer at Best Buy for 20% off the original price. If the original price of the desktop is $1500, how
Arturiano [62]

Answer:

So for this sale, you'll save $300.00 on this item.

This means, the cost of the item to you is

$1500 - $300.00 = $1200.00.

3 0
3 years ago
A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses
Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

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

where, \bar X = average age of the random sample of horses with colic = 12 yrs

            \mu = average age of all horses seen at the veterinary clinic = 10 yrs

   \sigma = standard deviation of all horses coming to the veterinary clinic = 8 yrs

         n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(\bar X \geq 12)

   P(\bar X \geq 12) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{12-10}{\frac{8}{\sqrt{60} } } ) = P(Z \geq 1.94) = 1 - P(Z < 1.94)

                                                 = 1 - 0.97381 = 0.0262

Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

4 0
3 years ago
The amount of water dispensed by a water dispenser is normally distributed,
Contact [7]

Answer:

The correct answer is - C.  11.45 ounces to 11.75 ounces.

Step-by-step explanation:

According to the empirical rule of the distribution for 68% falls under the normal curve falls within 1 standard deviation of the mean.

That is:

μ±δ

From the given information, the mean is

μ = 11.60

and the standard deviation is

δ = 0.15

We substitute the given parameters to obtain;

11.60±0.15

11.75 and 11.45

This means the lower limit is

11.45

and the upper limit is

11.75

3 0
3 years ago
Simplify this expression: cos t(sec t − cos t)
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\cos t \, (\sec t - \cos t)
= \cos t \, (\frac{1}{\cos t} - \cos t)
= 1 - \cos^2 t
= \bf sin^2 t
5 0
3 years ago
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