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Nataly [62]
3 years ago
8

You have a bag with 30 fireballs and another bag with 42 Jolly Ranchers.For a party , you want to repackage the candy into small

er bags,and you want each bag to have the same number of fireballs and jolly rancher’s as every other bag . How many bags will you be able to make?
Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
4 0

Answer:

6 bags 5 fire balls and 7 jolly ranchers in one bad

Step-by-step explanation:

Ok so you got 30 fire balls 42 ranchers what you can do is see like wow more ranchers in a bag we know that hmm 42/6 is ya know 7 and that 6 does it go into 30 yeah it does 30/6=5 so you got 5 fire balls and 7 jolly ranchers is each bag with 6 bags

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The cost C (in dollars) for the care and maintenance of a horse and carriage is C=15x+2000 where x is the number of rides.
MissTica

Answer: 100 rides are needed to break even.


Step-by-step explanation:

We have given the cost function

C(x)=15x+2000 where x is the number of rides.

And rides cost 35$

⇒ revenue function would be 35 times x

i.e.R(x)=35 x , where x is the number of rides.

Break even point of a firm occurs when at a certain point x the total cost equals to the total revenue.

i.e. at break even point

total revenue=total cost

⇒35x=15x+2000

⇒35-15x=2000[subtract 15x from both sides]

⇒20x=2000 [simplify]

⇒x=2000/20[dividing both sides with 20]

⇒x=100

∴ 100  rides are needed to break even.

6 0
3 years ago
If you understand may you please help me?​
Darina [25.2K]

Answer:

Step-by-step explanation:

3 0
3 years ago
A garden store has the following miscellaneous bulbs in a basket:
ArbitrLikvidat [17]
Okay. Since we know that the customer before them bought one of each type of bulb, we can subtract one bulb from each type so now, here is what the store has:

5 amaryllis

6 daffodils

3 lilies

2 tulips

Next, let’s find the probability of the customer picking an amaryllis.

We know there are 5 amaryllis, so we can put that as the numerator.

Next, we have to add up all the flowers to get the denominator.

5+6+3+2

I got 16.

So now we have 5/16

So, that is our answer! 5/16, or as a decimal 0.3125, or as a percentage 31.25%.

Hope this helps! Comment if you have any questions! :)
4 0
3 years ago
Which statement best explains the relationship between numbers divisible by 5 and 10
densk [106]

0 because you can divide them ie 0 is divisible with 5 and 10

4 0
3 years ago
The ideal size of a first-year class at a particular college is 150 students. The college, knowing from past experiences that on
katen-ka-za [31]

Answer:

6.18% probability that more than 150 first-year students attend this college.

Step-by-step explanation:

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

For each item selected, there are only two possible outcomes. Either it is defective, or it is not. This means that we use concepts of the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

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 = 450, p = 0.3

So

\mu = E(X) = np = 450*0.3 = 135

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{450*0.3*0.7} = 9.72

Approximate the probability that more than 150 first-year students attend this college.

This is 1 subtracted by the pvalue of Z when X = 150. So

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

Z = \frac{150 - 135}{9.72}

Z = 1.54

Z = 1.54 has a pvalue of 0.9382

1 - 0.9382 = 0.0618

6.18% probability that more than 150 first-year students attend this college.

7 0
4 years ago
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