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AVprozaik [17]
3 years ago
7

A box contains 3 plain pencils and 7 pens. A second box contains 9 color pencils and 3 crayons. One item from each box is chosen

at random. What is the probability that a pen from the first box and a crayon from the second box are selected? Write your answer as a fraction in simplest form.
Mathematics
1 answer:
Anastasy [175]3 years ago
3 0
P(P then C)=(7/10)(3/12)

P(PthenC)=21/120

P(PthenC)=7/40
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Suppose you like to keep a jar of change on your desk. Currently, the jar contains the following: 22 Pennies 27 Dimes 9 Nickels
kap26 [50]

Answer:

P(Penny\ and\ Dime) = \frac{9}{116}

Step-by-step explanation:

Given

Pennies = 22

Dimes = 27

Nickels = 9

Quarters = 30

Required

P(Penny\ and\ Dime)

This is calculated as:

P(Penny\ and\ Dime) = P(Penny) * P(Dime)

Since it is a selection without replacement, the computation is:

P(Penny\ and\ Dime) = \frac{Penny}{Total} * \frac{Dime}{Total-1}

So, we have:

P(Penny\ and\ Dime) = \frac{22}{22+27+9+30} * \frac{Dime}{22+27+9+30-1}

P(Penny\ and\ Dime) = \frac{22}{88} * \frac{27}{87}

P(Penny\ and\ Dime) = \frac{1}{4} * \frac{9}{29}

P(Penny\ and\ Dime) = \frac{9}{116}

5 0
3 years ago
At the local pet store, zebra fish cost $1.80 each and neon tetras cost $2.10 each. If Ernesto bought 16 fish for a total cost o
madreJ [45]
<span>zebra fish a then tetras fish (11 – a)
2.2a + 1.85(11 – a) = 22.80
2.2a + 20.35 – 1.85a = 22.80
0.35a = 2.45
a = 7 zebra and 11 – 7 = 4 neon fish

</span>
3 0
3 years ago
how many pounds of rock will you need to fill an area 25 ft by 35 ft to a depth of 2 in. If the rock weighs 1050 lb cubic yard?
Eduardwww [97]

The volume of the area is Length x width x height:

25 x 35 x 2/12 = 145.83 cubic feet.

1 cubic foot = 0.037 cubic yards:

145.83 x 0.037 = 5.40 cubic yards.

5.4 cubic yards x 1050 lbs/ cubic yard = 5,665.6 pounds.

Note:

Rounding and conversion may change the answer slightly, but I believe I rounded everything as it should be.

6 0
3 years ago
on the asymptotic behavior of the sample estimates of eigenvalues and eigenvectors of covariance matrices
skelet666 [1.2K]

A complex mathematical topic, the asymptotic behavior of sequences of random variables, or the behavior of indefinitely long sequences of random variables, has significant ramifications for the statistical analysis of data from large samples.

The asymptotic behavior of the sample estimators of the eigenvalues and eigenvectors of covariance matrices is examined in this claim. This work focuses on limited sample size scenarios where the number of accessible observations is comparable in magnitude to the observation dimension rather than usual high sample-size asymptotic .

Under the presumption that both the sample size and the observation dimension go to infinity while their quotient converges to a positive value, the asymptotic behavior of the conventional sample estimates is examined using methods from random matrix theory.

Closed form asymptotic expressions of these estimators are obtained, demonstrating the inconsistency of the conventional sample estimators in these asymptotic conditions, assuming that an asymptotic eigenvalue splitting condition is satisfied.

To learn more about asymptotic behavior visit:brainly.com/question/17767511

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4 0
1 year ago
Which is the equation of a line with a slope of -2 and a y-intercept of -5?
kogti [31]

Answer:

y = -2x - 5

Step-by-step explanation:

The given slope, or m, is -2 and the given y-intercept, or b, is -5.

The y-intercept equation is y = mx + b, so by knowing this equation all you need to do is plug in the given slope and y-intercept

y = -2x - 5

5 0
3 years ago
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