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I need help please I don’t know what to it is due today

Answers

Answer 1

Answer:

Answer:

12 blue necklaces + 12 red necklaces = 24 necklaces. He will have 1 blue bead left over and 1 red bead left over.

Step-by-step explanation:

37/3 = 12.333 or 12 r1

25/2= 12.5 or 12 r1

1+1=2

Answer 2

Answer: 1 red and 1 blue bead left over

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Gibson (1986) asked a sample of college students to complete a self-esteem scale on which the midpoint of the scale was the score 108. He found that the average self-esteem score for the samples well above the actual midpoint of the scale. Given that the standard deviation of self esteem scores was 28.15, what would be the score for a sample participant whose ser esteem score was a. 101.67 b. -0.23 c. -0.87 d.0 e. -1.19 f. 0.97

How did you do this question?

Answers

Answer:

R = ∞

I = (-∞, ∞)

Step-by-step explanation:

Use the ratio test:

lim(n→∞)│aₙ₊₁ / aₙ│

lim(n→∞)│[xⁿ⁺⁶ / (2(n+1)!)] / [xⁿ⁺⁵ / (2n!]│

lim(n→∞)│[xⁿ⁺⁶ / (2(n+1)!)] × (2n! / xⁿ⁺⁵)│

lim(n→∞)│x 2n! / (2(n+1)!)│

lim(n→∞)│n! / (n+1)!││x│

lim(n→∞) (1 / (n+1))│x│

The series converges if the limit is less than 1.

The limit is always less than 1, so the radius of convergence is infinite.

So the interval of convergence is (-∞, ∞).

Suppose that 2121% of all births in a certain region take place by Caesarian section each year. a. In a random sample of 500500 births, how many, on average, will take place by Caesarian section? b. What is the standard deviation of the number of Caesarian section births in a sample of 500500 births? c. Use your answers to parts a and b to form an interval that is likely to contain the number of Caesarian section births in a sample of 500500 births.

Answers

Answer:

a) $E(X) = np=500*0.21= 105$

b) $Sd(X) = √(82.95)= 9.108$

c) Assuming a the normality assumption we will have within 2 deviations from the mean most of the data from the distribution and the interval for this case would be:

$\mu -2\sigma = 105-2*9.108=86.785$

$\mu +2\sigma = 105+2*9.108=123.215$

So we expect about 86 and 123 most of the numbers of Caesarian section births

Step-by-step explanation:

For this case we can define the random variable X as the number of births in the Caesarian section and from the data given we know that the distribution of X is:

$X \sim Binom (n = 500, p=0.21$

Part a

The expected value for this distribution is given by:

$E(X) = np=500*0.21= 105$

Part b

The variance is given by:

$Var(X) = np(1-p) = 500*0.21*(1-0.21)= 82.95$

And the deviation would be:

$Sd(X) = √(82.95)= 9.108$

Part c

Assuming a the normality assumption we will have within 2 deviations from the mean most of the data from the distribution and the interval for this case would be:

$\mu -2\sigma = 105-2*9.108=86.785$

$\mu +2\sigma = 105+2*9.108=123.215$

So we expect about 86 and 123 most of the numbers of Caesarian section births

I NEED HELP ASAP PLEASE HELP

Answers

Answer: C

Step-by-step explanation:

Answer:

Answer: C

Step-by-step explanation:

When looking at the whole number, You always want to look at the tenths place to see of it's 5 or higher.

Eric jog 3 1/4 mile on Monday, 5 5/8 miles on Tuesday, and 8 miles on Wednesday. Suppose he continues the pattern for the remainder of the week. How far will Ereck jog on Friday

Answers

Answer:

12 3/4 miles

Step-by-step explanation:

Eric is increasing his distance by 2 3/8 miles per day. In 2 more days, his distance will be 2×(2 3/8) = 4 3/4 miles more than on Wednesday.

... 8 + 4 3/4 = 12 3/4 . . . . miles on Friday

_____

Rate of increase

The difference between Tuesday and Monday is ...

... 5 5/8 - 3 2/8 = 2 3/8

The difference between Wednesday and Tuesday is ...

... 8 - 5 5/8 = 2 3/8

Thus, it appears that Eric is jogging 2 3/8 mile more each day than the day before.

Subtract breaking apart 143-79

Answers

Step-by-step explanation:

A two-digit number, such as 79, contains a tens place and a ones place.

That is because 79 is also equal to 70 + 9.

The tens place is thus the 7, since since 7 × 10 = 70

and the ones place is the 9.

So lets Subtract breaking apart 143 - 79.

Break 79 down into its parts:

70 + 9.

Subtract 70 from the original number:

143 - 70 = 73

Subtract 9 from the answer:

73 - 9 = 64

Therefore, 64 is the final answer.

Henry buys a bag of 15 tangerines for $2.85.Find the unit price in dollars per tangerine.
If necessary, round your answer to the nearest cent.
Please help with this

Answers

Answer:

.19 cents per tangerine

Step-by-step explanation:

.19 x 15 = 2.85

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