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3 years ago

Francine uses \dfrac23

Mathematics

1 answer:

pashok25 [27]3 years ago

7 0

Answer:

$\frac{1}{2}$ cup of pineapple juice.

Step-by-step explanation:

We are told that Francine uses 2/3 cup of pineapple juice for every 1/3 cup of orange juice to make a smoothie. In order to get the number of cups of pineapple juice Francine will use for 1 cup of orange juice, we will apply the following expression;

2/3 cup of pineapple juice = 1/3 cups of orange juice

1 cup of pineapple juice = x cups of orange juice

Cross multiply

$\frac{2}{3}\times x = 1 \times \frac{1}{3}\\ \\\frac{2x}{3} = \frac{1}{3}\\ \\cross \ multiply\\2x \times 3 = 3\\$

divide both sides of the equation by 3

$\frac{2x \times 3}{3} =\frac{3}{3}\\ \\2x = 1\\$

divide both sides by 2

$\frac{2x}{2} = \frac{1}{2}\\ x = \frac{1}{2}\\$

Hence Francine uses $\frac{1}{2}$ number of cups of pineapple juice for 1 cup of orange juice.

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Anika is 7 years older than Kit. If Kit is k years old, how old is Anika?

zaharov [31]

Answer:

k+7

Step-by-step explanation:

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7 0

3 years ago

Suppose that 15% of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in t

hram777 [196]

Answer:

a. $\mu=15$

b. $\mu=7.8586 \ and \ \mu=22.1414$

c. Choice A- Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

Step-by-step explanation:

a.Binomial distribution is defined by the expression

$P(X=k)=C_k^n.p^k.(1-p)^{n-k}$

Let n be the number of trials, $n=100$

and p be the probability of success, $p=15\%$

The mean of a binomial distribution is the probability x sample size.

$\mu=np=100\times0.15=15$

b.Limits within which p is approximately 95%

sd of a binomial distribution is given as: $\sigma=\sqrt npq\\q=1-p$

Therefore, $\sigma=\sqrt(100\times0.015\times0.85)=3.5707$

Use the empirical rule to find the limits. From the rule, approximately 95% of the observations are within to standard deviations from mean.

$sd_1=>\mu-2\sigma=15-2\times3.3507=7.8586\\sd_2=>\mu-2\sigma=15+2\times3.3507=22.1414$

Hence, approximately 95% of the observations are within 7.8586 and 22.1414 (areas of infestation).

c. $x=45$ is not within the limits in b above (7.8586,22.1414). X=45 appears to be a large area of infestation. A.Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.