How many times larger is 6 × 1011 than 2 × 10-5?

Use the drop-down menus to complete the statements about factoring 14x2 + 6x – 7x – 3 by grouping.

Stella [2.4K]

Answer:1. GCF of the group ( 6 x - 3 ) is 3.

2. The common binomial factor is 2 x - 1.

3. The factored expression is: ( 2 x - 1 ) ( 7 x + 3 ).

Step-by-step explanation:

14 x² + 6 x - 7 x - 3 =

= ( 14 x² - 7 x ) + ( 6 x - 3 ) =

= 7 x ( 2 x - 1 ) + 3 ( 2 x - 1 ) =

= ( 2 x - 1 ) ( 7 x + 3 )

4 0

2 years ago

Read 2 more answers

A sample of 11 students using a Ti-89 calculator averaged 31.5 items identified, with a standard deviation of 8.35. A sample of

lina2011 [118]

Answer:

We reject H₀, with CI = 90 % we can conclude that the mean numbers of times identified with Ti-84 does not exceed that of the Ti-89 by more than 1,80

Step-by-step explanation:

Ti-89 Calculator

Sample mean x = 31,5

Sample standard deviation s₂ = 8,35

Sample size n₂ = 11

Ti-84 Calculator

Sample mean y = 46,2

Sample standard deviation s₁ = 9,99

Sample size n₁ = 12

t(s) = ( y - x - d ) / √s₁²/n₁ + s₂²/n₂

t(s) = 12,9 / √(99,8/12) + (69,72/11)

t(s) = 12,9 / √8,32 + 6,34

t(s) = 12,9 / 3,83

t(s) = 3,37

Test Hypothesis

Null Hypothesis H₀ y - x > 1,80

Altenative Hypothesis Hₐ y - x ≤ 1,80

We have a t(s) = 3,37

We need to compare with t(c) critical value for

t(c) α; n₁ +n₂-2 df = 12 +11 -2 df = 21

If we choose CI = 95 % then α = 5 % α = 0,05

From t-student table

t(c) = 1,72

t(s) = 3,37

t(s) >t(c)

t(s) is in the rejection region therefore we accept Hₐ with CI = 95 %

8 0

3 years ago

Someone help me PLEASE

musickatia [10]

Answer:

that looks hard im sorry cant do that

Step-by-step explanation:

5 0

3 years ago

Please,someone can help me this question!!!

Anvisha [2.4K]

Answer:

x=126°, y=234°

Step-by-step explanation:

B altrenates A

180-54=126, x=126°

D is 180°, 54+180=234°

3 0

3 years ago

The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books check

Verdich [7]

Answer:

$n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11$

So the answer for this case would be n=11 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma =150$ represent the population standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The confidence interval for this case is given by: (740, 920)

We can find the estimate for the mean and we got:

$\bar X = \frac{740+920}{2} = 830$

and the margin of error is given by :

$ME = \frac{920-740}{2}= 90$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =90 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got:

$n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11$

So the answer for this case would be n=11 rounded up to the nearest integer

5 0

2 years ago