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FinnZ [79.3K]
3 years ago
13

There are three green, three blue and three white marbles in a jar. what is the probability of drawing first a white marble and

then a blue marble, without replacing the first marble into the jar?
Mathematics
1 answer:
Anna11 [10]3 years ago
4 0
The probability of first drawing a white marble is 3/9. Once you've drawn it, the probability of drawing a blue marble is 3/8. So you multiply the fractions and get 9/72 as your answer. You can simplify to 1/8 if you'd like. Hope this helps!
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It costs you $10 to draw a sample size n=1 and measure the attribute of interest. you have a budget of $1,200
kicyunya [14]

<span>a.       </span>Do you have sufficient funds to estimate the population mean for the attribute of interest with a 95% confidence interval 4 units width? Assume that sd= 12

n= {[(Zalpha/2)^2]*[sd]^2}/ se^2

n= (1.96)^2*(12)^2/ (2)^2

n= 138.297 rounded up to 139

<span>There is not enough funds for this one since you’ll need 139 pieces and it costs 10 a piece, you’ll need 1390.</span>

b.      90% confidence interval

n= {[(Zalpha/2)^2]*[sd]^2}/ se^2

n= (1.645)^2*(12)^2/ (2)^2

n=98

There is enough funds since 98 pieces for 10 a piece is equal to 980.

6 0
4 years ago
The cost of 900 shares of stock selling for 57 5/8 is
Aleksandr [31]

Answer:

  $51,862.50

Step-by-step explanation:

... the product of the number of shares and the price:

  (900)($57 5/8) = $51,862.50

3 0
4 years ago
Read 2 more answers
Weights of cars passing over a bridge part 2<br>​
DaniilM [7]

Step-by-step explanation:

The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.

Given:

mean,  mu = 3550 lbs (hope I read the first five correctly, and it's not a six)

standard deviation, sigma = 870 lbs

weights are normally distributed, and assume large samples.

Probability to be estimated between W1=2800 and W2=4500 lbs.

Solution:

We calculate Z-scores for each of the limits in order to estimate probabilities from tables.

For W1 (lower limit),

Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069

From tables, P(Z<Z1) = 0.194325

For W2 (upper limit):

Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954

From tables, P(Z<Z2) = 0.862573

Therefore probability that weight is between W1 and W2 is

P( W1 < W < W2 )

= P(Z1 < Z < Z2)

= P(Z<Z2) - P(Z<Z1)

= 0.862573 - 0.194325

= 0.668248

= 0.67 (to the hundredth)

6 0
3 years ago
Translate the phrase into a numerical expression of One hundred divided by the quantity three times five
dimulka [17.4K]
\frac{100}{3*5}
4 0
4 years ago
The shaded sector of the circle shown above has an area of 18 pi square feet
Genrish500 [490]

Answer:

F

Step-by-step explanation:

Area of sector=pi*r^2*(theta/360) = pi*r^2*(45/360)

18*pi=pi*r^2*(1/8), r=12. Circumference is 24*pi

5 0
3 years ago
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