All categories

3 years ago

What is the ratio of 2 cups of flour for every 10 pancakes

Mathematics

1 answer:

julsineya [31]3 years ago

8 0

1:5 or 2:10 because if you divide 10 by 2 you get 5 and 2 divided by 2 it is 1

You might be interested in

A box contains 3 green marbles, 5 blue marbles, and 7 red marbles. Three marbles are selected at random from the box, one at a t

USPshnik [31]

Answer:

The probability is $P(K) = \frac{ 28 }{195}$

Step-by-step explanation:

From the question we are told that

The number of green marbles is $n_g = 3$

The number of red marbles is $n_b = 5$

The number of red marbles is $n_r = 7$

Generally the total number of marbles is mathematically represented as

$n_t = n_r + n_g + n_ b$

$n_t = 7 + 3 + 5$

$n_t = 5$

Generally total number of marbles that are not red is

$n_k = n_g + n_ b$

=> $n_k = 3 + 5$

=> $n_k = 8$

The probability of the first ball not being red is mathematically represented as

$P(r') = \frac{n_k}{n_t}$

=> $P(r') = \frac{ 8}{15}$

The probability of the second ball not being red is mathematically represented as

$P(r'') = \frac{n_k - 1}{n_t -1}$

=> $P(r'') = \frac{ 8 -1 }{15-1}$ (the subtraction is because the marbles where selected without replacement )

=> $P(r'') = \frac{ 7 }{14}$

The probability that the first two balls is not red is mathematically represented as

$P(R) = P(r') * P(r'')$

=> $P(R) = \frac{ 8}{15} * \frac{ 7 }{14}$

=> $P(R) = \frac{ 8 }{30}$

The probability of the third ball being red is mathematically represented as

$P(r) = \frac{n_r}{ n_t -2}$ (the subtraction is because the marbles where selected without replacement )

$P(r) = \frac{7}{ 15 -2}$

=> $P(r) = \frac{7}{ 13}$

Generally the probability of the first two marble not being red and the third marble being red is mathematically represented as

$P(K) = P(R) * P(r)$

$P(K) = \frac{ 8 }{30} * \frac{7}{ 13}$

=> $P(K) = \frac{ 28 }{195}$

6 0

3 years ago

A manufacturer of matches randomly and independently puts 23 matches in each box of matches produced. The company knows that one

d1i1m1o1n [39]

Answer:

0.9855 or 98.55%.

Step-by-step explanation:

The probability of each individual match being flawed is p = 0.008. The probability that a matchbox will have one or fewer matches with a flaw is the same as the probability of a matchbox having exactly one or exactly zero matches with a flaw:

$P(X\leq 1)=P(X=0)+P(X=1)\\P(X\leq 1)=(1-p)^{23}+23*(1-p)^{23-1}*p\\P(X\leq 1)=(1-0.008)^{23}+23*(1-0.008)^{23-1}*0.008\\P(X\leq 1)=0.8313+0.1542\\P(X\leq 1)=0.9855$