Ask question

Ask question

All categories

3 years ago

10

What is the gcf for 28 and 70

1 answer:

Lena [83]3 years ago

8 0

28 factors: 1,2,4,7,14,28
70 factors: 1,2,5,7,10,14,35,70
GCF : 14

You might be interested in

Drew is making a dart board.

Nezavi [6.7K]

Answer:

138.16 cm

Step-by-step explanation:

Recall that the formula for calculating the circumference is pi times diameter.

Circumference = 44 x 3.14 cm = 138.16 cm.

5 0

3 years ago

If two cards are drawn out of a deck WITH replacement, what's the probability that you draw a red card and then an Ace? ** Remem

katen-ka-za [31]

Answer:

Step-by-step explanation:

The probability of drawing a red card is 1/2, since half of the cards in a deck are red. The probability of drawing an ace from a deck is 1/13, since there is an equal chance to pick any of the 13 numbers.

P(red card and ace)= P(red card)*P(ace)=1/2 * 1/13=1/26

4 0

3 years ago

What number is 0.25% of 46?

cestrela7 [59]

Answer:

0.115

Step-by-step explanation:

First, multiply 0.25 x 46 to get 11.5

Then, multiply by 11.5 x 100 and get your answer.

You can find the percent of any number my doing this:

Multiplying the percent x the number then dividing by 100

hope this helps :)

3 0

3 years ago

Help is needed! Desperate

Bogdan [553]

Sorry I don't know that question.

5 0

3 years ago

Read 2 more answers

A geologist has collected 5 specimens of basaltic rock and 7 specimens of granite. The geologist instructs a laboratory assistan

MaRussiya [10]

The rocks are chosen without replacement, which means that the hypergeometric distribution is used to solve this question. First we get the parameters, and then we answer the questions. From this, we get that:

$E(X) = 5.25, Var(X) = 0.5966$
P(X < 6) = 0.9545
P(all specimens of one of the two types of rock are selected for analysis) = 0.2046.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

The mean and the variance are:

$\mu = \frac{nk}{N}$

$\sigma^2 = \frac{nk(N-k)(N-n)}{N^2(N-1)}$

We have that:

5 + 7 = 12 rocks, which means that $N = 12$

9 are chosen, which means that $n = 9$

7 are granite, which means that $k = 7$

Question a:

$E(X) = \mu = \frac{9\times7}{12} = 5.25$

$Var(X) = \sigma^2 = \frac{9\times7(12-7)(12-9)}{12^2(12-1)} = 0.5966$

Thus:

$E(X) = 5.25, Var(X) = 0.5966$

Question b:

Since there are only 5 specimens of basaltic rock, at least 9 - 5 = 4 specimens of granite are needed, which means that:

$P(X < 6) = P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 4) = h(4,12,9,7) = \frac{C_{7,4}*C_{5,5}}{C_{12,9}} = 0.1591$

$P(X = 5) = h(5,12,9,7) = \frac{C_{7,5}*C_{5,4}}{C_{12,9}} = 0.4773$

$P(X = 6) = h(6,12,9,7) = \frac{C_{7,6}*C_{5,3}}{C_{12,9}} = 0.3181$

Thus

$P(X < 6) = P(X = 4) + P(X = 5) + P(X = 6) = 0.1591 + 0.4773 + 0.3181 = 0.9545$

So P(X < 6) = 0.9545.

Question c:

5 of basaltic and 4 of granite: 0.1591 probability.

7 of granite is P(X = 7), in which

$P(X = 7) = h(7,12,9,7) = \frac{C_{7,7}*C_{5,2}}{C_{12,9}} = 0.0455$

0.1591 + 0.0455 = 0.2046, thus:

P(all specimens of one of the two types of rock are selected for analysis) = 0.2046.

A similar question is found at brainly.com/question/24008577

5 0

3 years ago