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valentinak56 [21]

3 years ago

13

Please help! This was due yesterday :(

1 answer:

Julli [10]3 years ago

6 0

Answer mon

Step-by-step explanation:

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stiv31 [10]

B, C, and D are the correct answers

3 0

2 years ago

The following parts refer to the letters: LAMEFIREALARM. Recall that "word" means distinguishable letter arrangements. (a) How m

larisa86 [58]

Answer:

Step-by-step explanation:

The word "LAMEFIREALARM" word consists of

2L , 2M, 2E , 2R , 3A , F and I

no of words with the above letters such that M's are not together

=Total No of words- M's are together

Total no of words $=\frac{13!}{4\times 2!\times 3!}=64,864,800$

When M's are together . considering 2 M's as one

therefore there are $\frac{12!}{3\times 2!\times 3!}=9,979,200$

No of ways $=\frac{13!}{4\times 2!\times 3!}-\frac{12!}{3\times 2!\times 3!}=54,885,600$

(b)No of ways such that M's are separated by at least 2 letters

at least 2 letter means 2 letter, 3 letter ......11 letters

So we have to subtract no of ways where there are 1 letter in between M's from total no of ways where 2 M's are not next to each other

No of ways in which there is 1 letter between 2 M's

This can be done by considering 11 cases

In first case Place first M in Starting Position and 2 M on third place

Second case place First M in 2 nd Position and second M on 4 th place

Similarly For 11th case

Place first M in 11th place and second M on 13th place

Total ways

$=\frac{11\times 11!}{3!\times 2!\times 2!\times 2!}$

So , the total number of arrangements of the letters of the word LAMEFIREALARM where the two M's are separated by atleast two letters is

$=\frac{13!}{4\times 2!\times 3!}-\frac{12!}{3\times 2!\times 3!}-\frac{11\times 11!}{3!\times 2!\times 2!\times 2!}$

$=45738000$

7 0

3 years ago

Write 8 as a product of primes.<br> Use index notation

dimaraw [331]

Answer:

2³

Step-by-step explanation:

8 = 2 × 2 × 2 = 2³

4 0

2 years ago

When there is a wedge between private costs and social costs, which of the following must be present?

Bogdan [553]

The negative externalities must be present.

4 0

2 years ago

The following results come from two independent random samples taken of two populations.

photoshop1234 [79]

Answer:

$(a)\ \bar x_1 - \bar x_2 = 2.0$

$(b)\ CI =(1.0542,2.9458)$

$(c)\ CI = (0.8730,2.1270)$

Step-by-step explanation:

Given

$n_1 = 60$ $n_2 = 35$

$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

$\sigma_1 = 2.1$ $\sigma_2 = 3$

Solving (a): Point estimate of difference of mean

This is calculated as: $\bar x_1 - \bar x_2$

$\bar x_1 - \bar x_2 = 13.6 - 11.6$

$\bar x_1 - \bar x_2 = 2.0$

Solving (b): 90% confidence interval

We have:

$c = 90\%$

$c = 0.90$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.90$

$\alpha = 0.10$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.10/2}$

$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

$\alpha = 0.05$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

$z_{\alpha/2} = z_{0.025} =1.96$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.96 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.96* \sqrt{0.3306}$

$2.0 \± 1.1270$

Split

$(2.0 - 1.1270) \to (2.0 + 1.1270)$

$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

$CI = (0.8730,2.1270)$

8 0

3 years ago