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zlopas [31]
2 years ago
13

5 pencils are in a bag.There are 3 times as many pens than pencils in the same bag. How many pens are in the bag

Mathematics
2 answers:
Mkey [24]2 years ago
7 0
Ur answer is 15 because is three times as much pencils in the bag so 5×3=15
Darya [45]2 years ago
6 0
There are 15 pens in the bag
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Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a
Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }

Z*(Sigma/\sqrt{n} )= (X[bar]-Mu)

(Sigma/\sqrt{n} )= \frac{(X[bar]-Mu)}{Z}

Sigma= \frac{(X[bar]-Mu)}{Z}*\sqrt{n}

Sigma*(\frac{Z}{(X[bar]-Mu)})= \sqrt{n}

n = (Sigma*(\frac{Z}{(X[bar]-Mu)}))^2

n = (16*(\frac{1.2}{(270-264)}))^2

n= 10.24 ≅ 11 pregnant women.

I hope it helps!

6 0
2 years ago
Find the simple interest paid to the nearest cent for each loan, interest rate, and time. $500, 12%, 18 months
iris [78.8K]

Answer: The interest is $60


Step-by-step explanation:

500*112%=$560

560/18=$31.11

$60 in interest is paid, or an additional $3.33 a month (Bringing the monthly loan payment to a grand total of $31.11)

4 0
2 years ago
Read 2 more answers
A company provides a monthly pension to its employees. A person retiring at 62 retires and receives a full monthly pension. If t
IRISSAK [1]

Answer:

  • <em><u>The reduction is 8.6%</u></em>

Explanation:

Call F the full monthly pension of a person retiring at 62.

If a person continues to work the pension grows at a rate of 6% per year, compounded monthly, so use the compounded growing formula:

  • Pension=F(1+r/12)^{12t}

Where r = 6 / 100 = 0.06, and t = number of years after retirement.

<u>For retirement at 65.5</u>:

  • t = 65.5 - 62 = 3.5

  • Pension=F(1+0.06/12)^{12\times 3.5}=1.233F

<u>For retirement at 67</u>:

  • t = 67 - 62 = 5

  • Pension=F(1+0.06/12)^{12\times 5}=1.349F

<u>Percent reduction of people who retire at 65.5 compared to what they would receive at 67</u>:

  • (1.349F-1.233F)\times 100/(1.349F)=8.6\%
8 0
3 years ago
The answer or what measurements I need to multiply?
Zigmanuir [339]
Spilt into two figures
first one = 10 x 8 x 14 = 1120
second one = 10 x 10 x 8 = 800
Total volume = 1120 + 800 = 1920 cubic cm
6 0
3 years ago
What is the varible of -10 = xy +z for x
Verdich [7]

The variable is y

Z is a constant.

4 0
3 years ago
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