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3 years ago

-81, 108, -144, 192, ...

Mathematics

1 answer:

noname [10]3 years ago

8 0

Answer:

Step-by-step explanation:

The terms form a geometric sequence with common ratio r

r = $\frac{a_{2} }{a_{1} }$ = $\frac{a_{3} }{a_{2} }$ = ......

r = $\frac{108}{-81}$ = $\frac{-144}{108}$ = - $\frac{4}{3}$

The n th term formula of a geometric sequence is

$a_{n}$ = a $(r)^{n-1}$ ← a is the first term

here a = - 81, hence

f(x) = - 81 $(-\frac{4}{3}) ^{x-1}$ → C

You might be interested in

((-2^2)(-1^-3))•((-2^-3)(-1^5))^-2

Vitek1552 [10]

Answer:

256

Step-by-step explanation:

A calculator works well for this.

_____

None of the minus signs are subject to the exponents (because they are not in parentheses, as (-1)^5, for example. Since there are an even number of them in the product, their product is +1 and they can be ignored.

1 to any power is still 1, so the factors (1^n) can be ignored.

After you ignore all of the things that can be ignored, your problem simplifies to ...

(2^2)(2^-3)^-2

The rules of exponents applicable to this are ...

(a^b)^c = a^(b·c)

(a^b)(a^c) = a^(b+c)

Then your product simplifies to ...

(2^2)(2^((-3)(-2)) = (2^2)(2^6)

= 2^(2+6)

= 2^8 = 256

3 0

3 years ago

a math teacher claims that she has developed a review course that increases the score of students on the math portion of a colle

madam [21]

Answer:

$H_0: \mu\leq514\\\\H_1: \mu>514$

B. Z=2.255. P=0.01207.

C. Although it is not big difference, it is an improvement that has evidence. The scores are expected to be higher in average than without the review course.

D. P(z>0.94)=0.1736

Step-by-step explanation:

<em>A. state the null and alternative hypotheses.</em>

The null hypothesis states that the review course has no effect, so the scores are still the same. The alternative hypothesis states that the review course increase the score.

$H_0: \mu\leq514\\\\H_1: \mu>514$

B. test the hypothesis at the a=.10 level of confidence. is a mean math score of 520 significantly higher than 514? find the test statistic, find P-value. is the sample statistic significantly higher?

The test statistic Z can be calculated as

$Z=\frac{M-\mu}{s/\sqrt{N}} =\frac{520-514}{119/\sqrt{2000}}=\frac{6}{2.661}=2.255$

The P-value of z=2.255 is P=0.01207.

The P-value is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.

Then we can conclude that the score of 520 is significantly higher than 514, in this case, specially because the big sample size.

C. do you think that a mean math score of 520 vs 514 will affect the decision of a school admissions adminstrator? in other words does the increase in the score have any practical significance?

Although it is not big difference, it is an improvement that has evidence. The scores are expected to be higher in average than without the review course.

D. test the hypothesis at the a=.10 level of confidence with n=350 students. assume that the sample mean is still 520 and the sample standard deviation is still 119. is a sample mean of 520 significantly more than 514? find test statistic, find p value, is the sample mean statisically significantly higher? what do you conclude about the impact of large samples on the p-value?

In this case, the z-value is

$Z=\frac{520-514}{s/\sqrt{n}} =\frac{6}{119/\sqrt{350}} =\frac{6}{6.36} =0.94\\\\P(z>0.94)=0.1736>\alpha$

In this case, the P-value is greater than the significance level, so there is no evidence that the review course is increasing the scores.

The sample size gives robustness to the results of the sample. A large sample is more representative of the population than a smaller sample. A little difference, as 520 from 514, but in a big sample leads to a more strong evidence of a change in the mean score.

Large samples give more extreme values of z, so P-values that are smaller and therefore tend to be smaller than the significance level.

8 0

3 years ago

Find The surface area of a cone with a radius of 7 and slant hight of 9

liubo4ka [24]

Answer:14

Step-by-step explanation:

4 0

3 years ago

X y

IRISSAK [1]

Answer: let me check