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Elena L [17]
3 years ago
8

In writing a word phrase from an equation, there is only one correct way to do it.

Mathematics
2 answers:
Nana76 [90]3 years ago
7 0

Answer:

False

Step-by-step explanation:

GalinKa [24]3 years ago
5 0
False it’s false dk what else to say
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kwasi thought of a number, multiplied it by7/2 and added 16 to the results.if the final answer was 30.what number did he think o
ziro4ka [17]

Answer:

4

Step-by-step explanation:

  1. represent the number he thought of with x

\frac{7}{2} x + 16 = 30

  1. multiply through with the LCM which is 2
  2. 2 \times  \frac{7}{2}x + 16 \times 2 = 30 \times 2
  3. 7x + 32 = 60
  4. 7x = 60 - 32
  5. 7x = 28
  6. \frac{7x}{7}  =  \frac{28}{7}
  7. x = 4
7 0
3 years ago
g If a hypothesis test with a significance level of α rejects H0: μ1=μ2 in favor of Ha: μ1≠μ2, then thecorresponding (1- α)% con
Effectus [21]

Answer:

The statement, (1- <em>α</em>)% confidence interval for (μ₁ - μ₂) does not contain zero is TRUE.

Step-by-step explanation:

The hypothesis for a test is defined as follows:

<em>H</em>₀: μ₁ = μ₂ vs. <em>H</em>ₐ: μ₁ ≠ μ₂

It is provided that the test was rejected st the significance level <em>α</em>%.

If a decision is to made using the confidence interval the conditions are:

If the null hypothesis value is not included in the (1 - <em>α</em>)% confidence interval then the null hypothesis will be rejected and vice versa.

In this case the null hypothesis value is:

<em>H</em>₀: μ₁ - μ₂ = 0.

If the value 0 is not included in the (1 - <em>α</em>)% confidence interval for the difference between two means, then the null hypothesis will be rejected.

Thus the statement, (1- <em>α</em>)% confidence interval for (μ1- μ2) does not contain zero is TRUE.

5 0
3 years ago
M divided by 6 = 21 so what does m equal
loris [4]

Answer:

Step-by-step explanation:

M equals 3.5

6 0
3 years ago
(5p^2 - 3) + (2p^2 - 3p^3)
jeyben [28]
(5p² - 3) - (2p² - 3p³)
(5p² - 2p²) - 3 + 9p³
3p² - 3 + 9p³
4 0
3 years ago
If (x^2 - 1) is a factor of ax^4 + bx^3 + cx^2 + dx + e, show that a + c + e = b + d = 0
alisha [4.7K]
x^2-1=x^2-1^2=(x-1)(x+1)

If x²-1 is a factor of the polynomial, both x-1 and x+1 are factors of it.

According to the remainder theorem, if a binomial x-a is a factor of a polynomial p(x), then p(a)=0.

If x-1 and x+1 are factors of the polynomial p(x)=ax⁴+bx³+cx²+dx+e, then p(1)=0 and p(-1)=0.

p(1)=a \times 1^4+b \times 1^3 + c \times 1^2+ d \times 1+e \\&#10;p(-1)=a \times (-1)^4 + b \times (-1)^3 + c \times (-1)^2 + d \times (-1)+e \\ \\&#10;p(1)=a+b+c+d+e \\&#10;p(-1)=a-b+c-d+e \\ \\&#10;p(1)=0 \\&#10;p(-1)=0 \\ \\ \hbox{add both equations:} \\&#10;a+b+c+d+e=0 \\&#10;\underline{a-b+c-d+e=0} \\&#10;2a+2c+2e=0 \\&#10;2(a+c+e)=0 \\&#10;a+c+e=0 \\ \\&#10;\hbox{substitute 0 for a+c+e in the first equation:} \\&#10;a+b+c+d+e=0 \\&#10;(a+c+e)+b+d=0 \\&#10;0+b+d=0 \\&#10;b+d=0 \\ \\&#10;\boxed{a+c+e=b+d=0} \\&#10;\hbox{proved } \checkmark
8 0
3 years ago
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