Determine the sum of the following series. ∑n=1[infinity](4^n+8^n/13n)

spin [16.1K]

Two-third of the jar is filled with flower
Hope this helped

4 0

3 years ago

Jesse wants to paint the top of the table shown. What is the approximate area that she will paint? Circumference is 78 inches.

Salsk061 [2.6K]

Answer:

I need help with my question

Step-by-step explanation:

5 0

3 years ago

0.5m + 6.4 = 4.9 - 0.1m

BigorU [14]

Solve for m by simplifying both sides of the equation, then isolating the variable.
m = -2.5

5 0

3 years ago

in Colorado the state sales tax is 2.9%. if you buy a $2100 computer in Colorado, how much state tax will you pay?

Shtirlitz [24]

Answer:

$60.9

Step-by-step explanation:

Tax = 2.9% of Cost to be bought

= 2.9 / 100 × 2100

= $60.9

6 0

3 years ago

When a scientist conducted a genetics experiments with peas, one sample of offspring consisted of 943 peas, with 717 of them hav

pshichka [43]

Using the normal approximation to the binomial distribution, it is found that:

a) 0.242 = 24.2% probability of getting 717 or more peas with red flowers.

b) Since Z < 2, 717 peas with red flowers is not significantly high.

c) Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

For each pea, there are only two possible outcomes. Either they have a red flower, or they do not. The probability of a pea having a red flower is independent of any other pea, which means that the binomial distribution is used to solve this question.

Binomial distribution:

Probability of x successes on n trials, with p probability.

Normal distribution:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

If Z > 2, the result is considered <u>significantly high</u>.

If $np \geq 10$ and $n(1-p) \geq 10$ , the binomial distribution can be approximated to the normal with:

$\mu = np$

$\sigma = \sqrt{np(1-p)}$

In this problem:

943 peas, thus, $n = 943$
3/4 probability of being red, thus $p = \frac{3}{4} = 0.75$ .

Applying the approximation:

$\mu = np = 943(0.75) = 707.25$

$\sigma = \sqrt{np(1-p)} = \sqrt{943(0.75)(0.25)} = 13.297$

Item a:

Using continuity correction, this probability is $P(X \geq 717 - 0.5) = P(X \geq 716.5)$ , which is <u>1 subtracted by the p-value of Z when X = 716.5</u>.

Then:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{716.5 - 707.25}{13.297}$

$Z = 0.7$

$Z = 0.7$ has a p-value of 0.758.

1 - 0.758 = 0.242

0.242 = 24.2% probability of getting 717 or more peas with red flowers.

Item b:

Since Z < 2, 717 peas with red flowers is not significantly high.

Item c:

Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

A similar problem is given at brainly.com/question/25212369

6 0

3 years ago