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Damm [24]
2 years ago
8

at a restaurant the ratio of kids meal sold to adults meals sold was 5:4 if there were 20 kids meals sold what is the combined a

mount of kids and adults meals sold
Mathematics
1 answer:
Darya [45]2 years ago
7 0
5x-\ kid's\ meal\\
4x-\ adult\ meal\\\\
5x=20\ \ \ | divide\ by\ 5\\\\
x=4\\\\
5x+4x=9x=9*4=36\\\\
Together\ they\ have\ 36\ meals.
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wel
There you go. I’m pretty sure this is right.

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1 year ago
at the same time a 15 foot pole casts a 7.5 foot shadow a nearby tree casts an 11 foot shadow how tall is the tree​
soldier1979 [14.2K]

answer: 22

15/7.5 = x/11

cross multiply

7.5x = 165

divide by 7.5 on both sides to get the x alone

7.5x/7.5 = 165/7.5

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the tree is 22 feet

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4 0
2 years ago
We have n = 100 many random variables Xi ’s, where the Xi ’s are independent and identically distributed Bernoulli random variab
777dan777 [17]

Answer:

(a) The distribution of X=\sum\limits^{n}_{i=1}{X_{i}} is a Binomial distribution.

(b) The sampling distribution of the sample mean will be approximately normal.

(c) The value of P(\bar X>0.50) is 0.50.

Step-by-step explanation:

It is provided that random variables X_{i} are independent and identically distributed Bernoulli random variables with <em>p</em> = 0.50.

The random sample selected is of size, <em>n</em> = 100.

(a)

Theorem:

Let X_{1},\ X_{2},\ X_{3},...\ X_{n} be independent Bernoulli random variables, each with parameter <em>p</em>, then the sum of of thee random variables, X=X_{1}+X_{2}+X_{3}...+X_{n} is a Binomial random variable with parameter <em>n</em> and <em>p</em>.

Thus, the distribution of X=\sum\limits^{n}_{i=1}{X_{i}} is a Binomial distribution.

(b)

According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.  

The sample size is large, i.e. <em>n</em> = 100 > 30.

So, the sampling distribution of the sample mean will be approximately normal.

The mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu=p=0.50

And the standard deviation of the distribution of sample mean is given by,

\sigma_{\bar x}=\sqrt{\frac{\sigma^{2}}{n}}=\sqrt{\frac{p(1-p)}{n}}=0.05

(c)

Compute the value of P(\bar X>0.50) as follows:

P(\bar X>0.50)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{0.50-0.50}{0.05})\\

                    =P(Z>0)\\=1-P(Z

*Use a <em>z</em>-table.

Thus, the value of P(\bar X>0.50) is 0.50.

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3 years ago
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dezoksy [38]
The question is unclear.
5 0
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ANTONII [103]

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