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Vikki [24]
3 years ago
9

Write a quadratic function f whose zeros are 4 and -5

Mathematics
1 answer:
pshichka [43]3 years ago
6 0

Answer:

x^2+x-20

Step-by-step explanation:

x = 4

x - 4 = 0

x = -5

x + 5 = -5

(x + 5)(x - 4)

x^2+x-20

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Consider the following food chain:
USPshnik [31]

Answer:

A

Step-by-step explanation:

the plant will increase because no grasshopper will be present to eat them without the grasshopper the mice cannot eat so the mice population will decrease....

6 0
3 years ago
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One team c 82.5% of the football players weigh more then 200 pounds if 40 men are on the team how many weigh more then 200 pound
valentinak56 [21]
The answer is 33 because 40*0.825 or 40*82.5% is 33 men.
7 0
3 years ago
Margaret bought a video game, her father gave her 0.2 of the money, her aunt gave 0.5 of the money and she saved the rest. She s
const2013 [10]

Answer:

if Im correct it should be $8.40

Step-by-step explanation:

why well if you think of breaking down by say 1/4 of 168 all you do is divide 4 which is the whole amount by 0.2 which is the amount the father gave her so 4/0.2= 20 the you take your whole 168 and divide by 20 which the whole number for 0.2 the your answer should be this 168/20= 8.4 which then i assumed it was $8.40 hopefully that is correct hehe

5 0
3 years ago
A recent estimate by a large distributor of gasoline claims that 60% of all cars stopping at their service stations chose unlead
Anton [14]

Answer:

We reject the null hypothesis and conclude that at least 2 proportions differ from the stated value.

Step-by-step explanation:

There are 3 types of gas listed in the question.

Thus;

n = 3

DF = n - 1

DF = 3 - 1

DF = 2

Let's state the hypotheses;

Null hypothesis; H0: P_regular = P_super unleaded = 20%; P_i leaded = 60%

Alternative hypothesis; Ha: At least 2 proportions differ from the stated value.

Observed values are;

Regular gas; O = 51

Unleaded gas; O = 261

Super Unleaded; O = 88

Total observed values = 51 + 261 + 88 = 400

We are told that super unleaded and regular were each selected 20% of the time and that unleaded gas was chosen 60% of the time.

Thus, expected values are;

Regular gas; E = 20% × 400 = 80

Unleaded gas; E = 60% × 400 = 240

Super Unleaded; E = 20% × 400 = 80

Formula for chi Square goodness of fit is;

X² = Σ[(O - E)²/E]

X² = (51 - 80)²/80) + (261 - 240)²/240) + (88 - 80)²/80)

X² = 13.15

From the chi Square distribution table attached and using; DF = 2 and X² = 13.15, we can trace the p-value to be approximately 0.001

Also from online p-value from chi Square calculator attached, we have p to be approximately 0.001 which is similar to what we got from the table.

Now, if we take the significance level to be 0.05, it means the p-value is less than it and thus we reject the null hypothesis and conclude that at least 2 proportions differ from the stated value.

4 0
3 years ago
Please help me...........​
Dominik [7]
I think it’s the 1st one
8 0
3 years ago
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