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sdas [7]
3 years ago
15

Kate earns $6 per hour babysitting. She made this table to show the relationship between the number of hours she works (x) and t

he amount of money she earns in dollars (y). All of the following coordinate pairs correspond to a graph of the table EXEPT_____.
(10, 60)

(12, 2)

(12, 72)

(4, 24)

Mathematics
2 answers:
pychu [463]3 years ago
7 0
It would be B, (12, 2)<span />
swat323 years ago
7 0
The answer is B= (12,2)
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The events A and B are independent such that P(A)=0.5 and P(B)=0.7.
pochemuha

Answer:

the answer to this is 0.15 #4

8 0
3 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.

8 0
3 years ago
Help please !!!!!!!!!!<br> I'll give brainly to whoever gets this correct
iris [78.8K]

The answer is B. It is shifted up by 6*1.5-2-2 with is 9-2-2=5.

5 0
2 years ago
Labrador Retriever weighs 48 kg after a diet and exercise program the dog weighs 43 kilograms to determine if this shows a perce
ExtremeBDS [4]

Answer:

percentage change in weight ≈ 10%

Step-by-step explanation:

The dog weighed 48 kg after a diet and after an exercise program the dog had a weight of 43 kg. This means the dog loss weight since the dog weight decreased from an initial value of 48 kg to 43 kg. The decrease in weight can be calculate as

decrease in weight = original weight - new weight

original weight  = 48 kg

new weight = 43 kg

decrease in weight = 48 - 43 = 5 kg

Since the weight decrease their will be a percentage decrease in weight.

% decrease = decrease in weight/original weight × 100

% decrease = 5/48 × 100

% decrease = 500/48

% decrease = 10. 42666666667

percentage change in weight ≈ 10%

5 0
3 years ago
The first side of a triangle measures 5 in. less than the second side, the third side is 3 in. more than the first side, and the
Mariulka [41]
X is the second side. Side 1 is s-5. side 3 is s-2 (s-5+3). They all must add up to 17. Well there is your answer right there. S-2
5 0
3 years ago
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