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hoa [83]
2 years ago
5

What is the probability of drawing a spade from a standard deck of cards or a 7?

Mathematics
1 answer:
Alik [6]2 years ago
6 0

The probability of drawing a spade is 13/52 = 1/4

Because every hand has 13 cards and there is 52 cards in a deck

The probability of drawing a 7 is 4/52 or 1/13

Because there are seven 7 cards in a deck and 52 cards.

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DESPERATE WILL GIVE BRAINLIST AND THANKS
Vitek1552 [10]

Answer:

21/22 = repeating

60/120 = terminating

56/72 = repeating

11/121 = repeating

Step-by-step explanation:

i just know. fight me.

3 0
3 years ago
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55 divided (4 plus 7) times 6 plus(15 divided by3
Ne4ueva [31]

55/(4+7) x 6+(15/3)= 35

because...

  1. 55/(4+7)=5
  2. 5 x 6=30
  3. 30+(15/3)=35
  4. So in total it equals to 35.
5 0
3 years ago
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Mrs. Hilt bought 15 books. Sh paid $11 for each book. She later sold all 15 books for $25 each. What is the difference between t
sweet [91]
Total amount of money paid
=11 x 15
=$165

total amount of money sold
=25 x 15
=$375

difference
= 375 - 165
=$210
4 0
3 years ago
What is 1 4/5 as a decimal
Mariulka [41]
Divide 4 by 5. You will get 0.8 . So, your answer will be 1.8 because you already have a whole number, you will add the decimal to the whole number and you would get 1.8 .
6 0
3 years ago
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A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed i
Hunter-Best [27]

Answer:

a)  σ/√n= 1.43 min

c) Margin of error 2.8028min

d) [30.1972; 35.8028]min

e) n=62 customers

Step-by-step explanation:

Hello!

The variable of interest is

X: Time a customer stays at a restaurant. (min)

A sample of 49 lunch customers was taken at a restaurant obtaining

X[bar]= 33 mi

The population standard deviation is known to be δ= 10min

a) and b)

There is no information about the distribution of the population, but we know that if the sample is large enough, n≥30, we can apply the central limit theorem and approximate the distribution of the sample mean to normal:

X[bar]≈N(μ;σ²/n)

Where μ is the population mean and σ²/n is the population variance of the sampling distribution.

The standard deviation of the mean is the square root of its variance:

√(σ²/n)= σ/√n= 10/√49= 10/7= 1.428≅ 1.43min

c)

The CI for the population mean has the general structure "Point estimator" ± "Margin of error"

Considering that we approximated the sampling distribution to normal and the standard deviation is known, the statistic to use to estimate the population mean is Z= (X[bar]-μ)/(σ/√n)≈N(0;1)

The formula for the interval is:

[X[bar]±Z_{1-\alpha /2}*(σ/√n)]

The margin of error of the 95% interval is:

Z_{1-\alpha /2}= Z_{1-0.025}= Z_{0.975}= 1.96

d= Z_{1-\alpha /2}*(σ/√n)= 1.96* 1.43= 2.8028

d)

[X[bar]±Z_{1-\alpha /2}*(σ/√n)]

[33±2.8028]

[30.1972; 35.8028]min

Using a confidence level of 95% you'd expect that the interval [30.1972; 35.8028]min contains the true average of time the customers spend at the restaurant.

e)

Considering the margin of error d=2.5min and the confidence level 95% you have to calculate the corresponding sample size to estimate the population mean. To do so you have to clear the value of n from the expression:

d= Z_{1-\alpha /2}*(σ/√n)

\frac{d}{Z_{1-\alpha /2}}= σ/√n

√n*(\frac{d}{Z_{1-\alpha /2}})= σ

√n= σ* (\frac{Z_{1-\alpha /2}}{d})

n=( σ* (\frac{Z_{1-\alpha /2}}{d}))²

n= (10*\frac{1.96}{2.5})²= 61.47≅ 62 customers

I hope this helps!

3 0
3 years ago
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