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icang [17]
3 years ago
12

What is the coefficient of the fifth term in a binomial that is raised to the sixth power?

Mathematics
2 answers:
Mila [183]3 years ago
7 0

Answer:

It is 15. Therefore, the coefficient's of the fifth term of binomial expression raised to the power six is 15

Vladimir79 [104]3 years ago
3 0
The answer is 15 :))
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8 0
3 years ago
Read 2 more answers
I am a two digit square number. the sum of my digits is 13. what Square number am I
skelet666 [1.2K]
49 Because 7x7 is 49 & 4+9=13
5 0
3 years ago
Please help!!! <br> I’m so bad at math.
marishachu [46]

Answer:

x = - 9 , 6

Step-by-step explanation:

The given function is  

h(x) = \frac{x - 8}{x^{2} + 3x - 54}

⇒ h(x) = \frac{x - 8}{(x - 6)(x + 9)}

Therefore, the value of h(x) will be undefined for the denominator equal to zero.

So, (x + 9)(x - 6) = 0

⇒ x = - 9 and x = 6

Therefore, the values of x which are not in domain of function h(x) are x = - 9 , 6. (Answer)

4 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
Solve pls brainliest
MatroZZZ [7]

Answer:

18 square feet

Step-by-step explanation:

hope this helps :)

5 0
3 years ago
Read 2 more answers
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