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3 years ago

PLEASE HELP!!!

1 answer:

4 0

Sinc you're looking for cost per unicorn you do total made by number of unicorns sold

4,480,000,000,000/14,000,000= 32,000

so that means each unicorn is 32,000$.

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Find the length of the missing side. It’s a written response

andre [41]

Using Pythagorean’s theorem, the side length is sqrt(89)

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3 years ago

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Given the Arithmetic sequence A1,A2,A3,A4A1,A2,A3,A4

Bad White [126]

The given sequence is
a₁ = 29
a₂ = 39
a₃ = 49
a₄ = 59

This sequence is an arithmetic sequence. Th first term is a₁ = 29, and the common difference is d= 10.

The n-th term is
$a_{n}=a_{1}+(n-1)d$
The 33-rd termis
a₃₃ = 29 + (33 - 1)*10
= 29 + 320
= 349

Answer: a₃₃ = 349

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3 years ago

a group of 48 employees is taking 11 cars to the annual company picnic if each car will be occupied by 4 or 5 employees what i'd

erastova [34]

3 cars will occupy 5 employees and 8 cars will occupy 4 employees.

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3 years ago

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A baseball team won 8 games and lost 6. If the team continues at this ratio, how many wins will the team have after playing 21 g

DIA [1.3K]

Answer:

12 wins and 9 losses.

Step-by-step explanation:

Because they are winning 57.14% of there games so if 8 out of 14=57.14% then that means if they would have still one at the exact pace then.

12 wins out of 21 would mean they would have one 57.14% of there games.

Hope this helps have a great afternoon:)

4 0

2 years ago

Suppose you take an SRS of size 1000 of Connecticut students eligible to take the SAT and find that 85% plan to take the SAT dur

nadezda [96]

Answer:

$\mu_{p}=p=0.81$

Step-by-step explanation:

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

So under the null hypothesis the mean for the population proportion is p

$\mu_{p}=p=0.81$

And the standard deviationis given by:

$\sigma_{p}=\sqrt{\frac{p_0(1-p_o)}{n}}=\sqrt{\frac{0.81(1-0.81)}{1000}}=0.0124$

7 0

4 years ago