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Keith_Richards [23]
3 years ago
8

A scientist measured the amount of rain that fell in a town during

Mathematics
1 answer:
allochka39001 [22]3 years ago
6 0

Answer:

1.47 mm

Step-by-step explanation:

In week 1 there was 2.6 mm

In week 4 there was 4.07

You want to find the difference of week 1 and 4 meaning you have to subtract week 4's value from week 1's to find how much more rain there was in week 4 than week 1

You would do 4.07-2.6

This is equal to 1.47

THEREFORE there was 1.47 mm more rain in week 4 than week 1

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2 years ago
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Mary buys 8 widgets for $40.00 she adds $1.00 in enhancements to each widget and sells them for $9.00 each. What is Mary's estim
Maslowich
The gross profit margin is calculated using the following rule:
gross profit margin = total profit / total sales

Now, we need to get the values of total profit and total sale:
total profit = <span>9*8-(40+8)=24$
total sales = 9*8 = 72$

Now, we will substitute in the above equation:
gross profit margin = 24/72 = 1/3 = 0.3333334
% = 0.33333334*100 = 33.3334%</span>
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3 years ago
Write the following number using Scientific Notation. 0.000000000089g PLEASE HELP FAST
konstantin123 [22]

8.9x10^-11

5 0
3 years ago
Determine the probability of selecting a two-child family with one boy and one girl assuming boys and girls are equally likely
Fittoniya [83]

Using the binomial distribution, it is found that there is a 0.5 = 50% probability of selecting a two-child family with one boy and one girl.

For each child, there are only two possible outcomes, either it is a boy, or it is a girl. The probability of a child being a boy or being a girl is independent of any other child, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • Two children, hence n = 2.
  • Equally as likely to be a boy or a girl, hence p = 0.5.

The probability of one of each is P(X = 1), hence:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{2,1}.(0.5)^{1}.(0.5)^{1} = 0.5

0.5 = 50% probability of selecting a two-child family with one boy and one girl.

A similar problem is given at brainly.com/question/24863377

3 0
2 years ago
Construct a 99% confidthence interval for the population mean .Assume the population has a normal distribution. A group of 19 ra
Kobotan [32]
<h2>Answer with explanation:</h2>

Confidence interval for mean, when population standard deviation is unknown:

\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}

, where \overline{x} = sample mean

n= sample size

s= sample standard deviation

t_{\alpha/2} = Critical t-value for n-1 degrees of freedom

We assume the population has a normal distribution.

Given, n= 19 , s= 3.8 , \overline{x}=22.4

\alpha=1-0.99=0.01

A) Critical t value for \alpha/2=0.005 and degree of 18 freedom

t_{\alpha/2} = 2.8784

B) Required confidence interval:

22.4\pm ( 2.8744)\dfrac{3.8}{\sqrt{19}}\\\\=22.4\pm2.5058\\\\=(22.4-2.5058,\ 22.4+2.5058)=(19.8942,\ 24.9058)\approx(19.9,\ 24.9)

Lower bound = 19.9 years

Uppen bound = 24.9 years

C) Interpretation: We are 99% confident that the true population mean of lies in (19.9, 24.9) .

5 0
3 years ago
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