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Rufina [12.5K]
3 years ago
9

Sheep:cows:pigs=12:3:2. There are 150 more sheep than pigs. How many cows has she got?

Mathematics
2 answers:
kramer3 years ago
7 0
Sheep (s) = pigs (p) +150
s=p+150
s=6p
6p=p+150
5p=150
p=30

there are 30 pigs
30/2=15
15*3=45

there are 45 cows

here's is a little bit so you know this is correct
30x6=180
180-30=150
there is a 150 difference between sheep and pigs.

Hope this helps :)
elena55 [62]3 years ago
3 0
Information Given: She has 12 groups of sheep, 3 groups of cows and 2 groups of pigs. Question asks: There are 150 more sheep than pigs. How many cows has she got?
They figured 150 more sheep than pigs from 2 pigs X 75 sheep=150 lambs of 12 groups of sheep.
150 lambs/3 groups of cows=50 cows.
She has 50 cows from 3 groups of cows.

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A rectangular flower garden in Samantha's backyard has 220 feet around its edge. The width of the garden is 50 feet. What is the
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Step-by-step explanation:

Given

The perimeter of the garden is 220\ ft

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3 years ago
What is 62/120 expressed as a percent rounded to the nearest whole number?
Afina-wow [57]

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52%

Step-by-step explanation:

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A door delivery florist wishes to estimate the proportion of people in his city that will purchase his flowers. Suppose the true
kvv77 [185]

Answer:

99.74% probability that the sample proportion will be less than 0.1

Step-by-step explanation:

I am going to use the binomial approximation to the normal 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)}.

In this problem, we have that:

n = 276, p = 0.06

So

\mu = E(X) = np = 276*0.06 = 16.56

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{276*0.06*0.94} = 3.9454

What is the probability that the sample proportion will be less than 0.1

This is the pvalue of Z when X = 0.1*276 = 27.6. So

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

Z = \frac{27.6 - 16.56}{3.9454}

Z = 2.8

Z = 2.8 has a pvalue of 0.9974

99.74% probability that the sample proportion will be less than 0.1

5 0
3 years ago
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