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3 years ago

PLEASE HELP ME PLEASE!

Mathematics

1 answer:

maria [59]3 years ago

4 0

They can because of zero property. If we set them equal to zero we get their roots which is 3 and -2. This is the same on the x axis which is goes through. We can mark these points.

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According to the rational root theorem, the

nydimaria [60]

Answer:

Step-by-step explanation:

Sorry this is a tough one and I don’t want to give you the wrong answers sorry again

6 0

2 years ago

Read 2 more answers

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Tomtit [17]

Answer:

Step-by-step explanation:

Number 5 is third option

and number 6 is second option

3 0

3 years ago

A cow can give 5 gallons of milk if it eats 2 pounds of grass. Express x in terms of y, where y is the number of gallons of milk

spayn [35]

Answer:

x = 2.5y

Step-by-step explanation:

If a cow can give 5 gallons of milk after eating 2 pounds of grass.

Let's consider 'x' is the number of pounds of grass it eats and 'y' is the number of gallons of milk the cow gives.

Hence,

5y = 2x

We will simplify it to get what x equal to y by dividing 2 on both side we find,

x = 2.5y

x is equal to 2.5 y

By eating 1 pound of grass cow can give 2.5 gallons of milk.

7 0

3 years ago

Suppose the scores on a test given to all juniors in a school district are normally distributed with a mean of 74 and a standard

ivann1987 [24]

Answer:

$\mu -\sigma = 66 , \mu +\sigma = 82$

$\mu -2\sigma = 58 , \mu +2\sigma = 90$

$\mu -3\sigma = 50 , \mu +3\sigma = 98$

And in the figure attached we see the limits with the percentages associated.

Step-by-step explanation:

For this case we know that the random variable of interest is the scores on a test given to all juniors in a school district follows a normal distribution with the following parameters:

$X \sim N(\mu= 74, \sigma =8)$

For this case we know from the empirical rule that within one deviation from the mean we have approximately 68.2% of the data, within 2 deviations from the mean we have 95% and within 3 deviation 99.7%

We can find the limits and we got:

$\mu -\sigma = 66 , \mu +\sigma = 82$

$\mu -2\sigma = 58 , \mu +2\sigma = 90$

$\mu -3\sigma = 50 , \mu +3\sigma = 98$

And in the figure attached we see the limits with the percentages associated.

7 0

3 years ago

The Tran family and the Green family each used their sprinklers last summer. The water output rate for the Tran family's sprinkl

ser-zykov [4K]

Answer:

Tran family's sprinkler was used for 20 hours

Green's family's sprinkler was used for 30 hours

Step-by-step explanation:

Let the hours for which Tran family's sprinkler used is x hours

water output rate for the Tran family's sprinkler = 35L per hour

water output from Tran family's sprinkler in x hours = 35*x L = 35x

Let the hours for which Green family's sprinkler used is y hours

water output rate for the Green family's sprinkler = 40L per hour

water output from Green family's sprinkler in x hours = 40*y L = 40y

Given

The families used their sprinklers for a combined total of 50 hours

thus

x + y = 50 -------------------equation 1

y = 50-x

total water output of 1900L

35x+40y = 1900 -------------------equation 1

using y = 50-x in equation 2, we have

35x + 40(50-x) = 1900

35x + 2000 - 40x = 1900

=> -5x = 1900 - 2000 = -100

=> x = -100/-5 = 20

y = 50-20 = 30

Thus,

Tran family's sprinkler was used for 20 hours

Green's family's sprinkler was used for 30 hours

4 0

3 years ago