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2 years ago

Please help this is for my son

Mathematics

2 answers:

Snezhnost [94]2 years ago

6 0

Total cookies both baked

$\\ \tt\longmapsto \dfrac{1}{3}+\dfrac{2}{5}$

$\\ \tt\longmapsto \dfrac{5+6}{15}=\dfrac{11}{15}$

Cookies left:-

$\\ \tt\longmapsto 30-11/15(30)=30-22=8$

Anuta_ua [19.1K]2 years ago

3 0

Answer:

8 cookies left to be baked

Step-by-step explanation:

Len

Len bakes 1/3 of 30 cookies

To find 1/3 of 30, divide 30 by 3 then multiply by 1:

$\implies \frac{30}{3} \times1=10 \times 1=10$

Therefore, Len bakes 10 cookies

Sallesh

Sallesh bakes 2/5 of 30 cookies

To find 2/5 of 30, divide 30 by 5 then multiply by 2:

$\implies \frac{30}{5} \times2=6 \times 2=12$

Therefore, Sallesh bakes 12 cookies

Cookies left to bake

To calculate how many cookies are left to bake, subtract the found number of cookies Len and Sallesh have baked from 30:

30 - 10 - 12 = 8

So there are 8 cookies left to be baked.

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Consider lines An and CD

vichka [17]

Answer:

slope of AB = 2

slope CD = 2

Step-by-step explanation:

the equation of a line is y = ax+ b where a is the slope, b is the y-intercept

let us take yB - yA

→ yB - yA = axB + b - axA - b

→ yB - yA = a(xB - xA)

→ a = (yB - yA) / (xB = xA) = (6 - 2) / (-5 - (-7)) = 2

same goes for line CD

5 0

3 years ago

A selective college would like to have an entering class of 950 students. Because not all students who are offered admission acc

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Answer:

a) The mean is 900 and the standard deviation is 15.

b) 100% probability that at least 800 students accept.

c) 0.05% probability that more than 950 will accept.

d) 94.84% probability that more than 950 will accept

Step-by-step explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .