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2 years ago

-. Fill in the real number property below that justifies each step in combining to linear binomials,

Mathematics

1 answer:

andrey2020 [161]2 years ago

6 0

Answer:

Idk

Step-by-step explanation:

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How many times larger

Zigmanuir [339]

2*10^9 i'm pretty sure it would be d.

3 0

2 years ago

Determine 6th term in the geometric sequence whose first term is 3 and whose common ratio is -4.

GREYUIT [131]

$a_1=3$
$a_2=-4a_1$
$a_3=-4a_2=(-4)^2a_1$
$\cdots$
$a_6=-4a_5=\cdots=(-4)^5a_1$

$\implies a_6=(-4)^5\cdot3=-3072$

6 0

3 years ago

PLEASE HELP PLEASE PLEASE

Phantasy [73]

-2(-2)^4 + 5(-2)^2 - 4

-2(-2)= 4. 4^4= 16

5(-2)= -10. -10^2= 100 - 4= 96

96 + 16= 112

Answer: 112

8 0

3 years ago

If 9 is subtracted from two thirds of a number , the result is 7. find the number.

ankoles [38]

So what you do is (2/3)x-9=7. Add 9 to both sides (2/3)x=16. Multiply by reciprocal on both sides 16 x (3/2). 16 x 3 = 48. 48/2=24

7 0

3 years ago

Read 2 more answers

The weights of 67 randomly selected axles were found to have a variance of 3.85. Construct the 80% confidence interval for the p

VLD [36.1K]

Answer:

$3.13$

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 67

Variance = 3.85

We have to find 80% confidence interval for the population variance of the weights.

Degree of freedom = 67 - 1 = 66

Level of significance = 0.2

Chi square critical value for lower tail =

$\chi^2_{1-\frac{\alpha}{2}}= 51.770$

Chi square critical value for upper tail =

$\chi^2_{\frac{\alpha}{2}}= 81.085$

80% confidence interval:

$\dfrac{(n-1)S^2}{\chi^2_{\frac{\alpha}{2}}} < \sigma^2 < \dfrac{(n-1)S^2}{\chi^2_{1-\frac{\alpha}{2}}}$

Putting values, we get,

$=\dfrac{(67-1)3.85}{81.085} < \sigma^2 < \dfrac{(67-1)3.85}{51.770}\\\\=3.13$

Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.

8 0

2 years ago