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Maurinko [17]
3 years ago
14

A salesperson had the following sales: $15.50, $18.98, s16.8, $14, $18.50, and $22. What was the average sale?

Mathematics
1 answer:
Jlenok [28]3 years ago
5 0

Answer:

$ 17.63

Step-by-step explanation:

Given,

The sales of the salesperson are,

$15.50, $18.98, $16.8, $14, $18.50, and $22,

Sum of sales = 15.50 + 18.98 + 16.8 + 14  + 18.50 + 22 = 105.78

Also, the number of sales = 6,

Thus, the average sale = \frac{\text{Sum of sales}}{\text{Number of sales}}

=\frac{105.78}{6}

= $ 17.63

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zepelin [54]
Jacob must now invest $10,712,72 to have $50,000 for the party.
3 0
3 years ago
The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra
Snezhnost [94]
Let X denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by X_1,\ldots,X_{36}, each independently and identically distributed with distribution X_i\sim\mathcal N(26,7.2).

You want to find

\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)

Recall that if X\sim\mathcal N(\mu,\sigma), then the sampling distribution \overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right) with n being the size of the sample.

Transforming to the standard normal distribution, you have

Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}

so that in this case,

Z=6\dfrac{\overline X-26}{7.2}

and the probability is equivalent to

\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)
=\mathbb P(Z>1.481)\approx0.0693
5 0
3 years ago
How do you solve powers of powers in advanced algebra
Anna35 [415]
Ex: 2^2^2

Basically, the base is multiplied by the first power however large the number. Then the new base is multiplied by the second power to get an even larger number.
Say you have a question that asks for the power of the power of 2. Or 2^2^2.

You would do the equation 2^2 first, and end up with 4. But since you have another power, you would have the answer as 4^2. Which would then equal 16. But another equation that is similar is: 2^2^2 = 2^4
The two equations would still get the same answer, but would look entirely different.
7 0
3 years ago
The length of time that an auditor spends reviewing an invoice is approximately normally distributed with a mean of 600 seconds
Bumek [7]
Answer: 11.5% 

Explanation:


Since 1 minute = 60 seconds, we multiply 12 minutes by 60 so that 12 minutes = 720 seconds. Thus, we're looking for a probability that the auditor will spend more than 720 seconds. 

Now, we get the z-score for 720 seconds by the following formula:

\text{z-score} =  \frac{x - \mu}{\sigma}

where 

t = \text{time for the auditor to finish his work } = 720 \text{ seconds}
\\ \mu = \text{average time for the auditor to finish his work } = 600 \text{ seconds}
\\ \sigma = \text{standard deviation } = 100 \text{ seconds}

So, the z-score of 720 seconds is given by:

\text{z-score} = \frac{x - \mu}{\sigma}
\\
\\ \text{z-score} = \frac{720 - 600}{100}
\\
\\ \boxed{\text{z-score} = 1.2}

Let

t = time for the auditor to finish his work
z = z-score of time t

Since the time is normally distributed, the probability for t > 720 is the same as the probability for z > 1.2. In terms of equation:

P(t \ \textgreater \  720) 
\\ = P(z \ \textgreater \  1.2)
\\ = 1 - P(z \leq 1.2)
\\ = 1 - 0.885
\\  \boxed{P(t \ \textgreater \  720)  = 0.115}

Hence, there is 11.5% chance that the auditor will spend more than 12 minutes in an invoice. 
8 0
3 years ago
Jason knows that the equation to calculate the period of a simple pendulum is , where T is the period, L is the length of the ro
Brrunno [24]

<u>Answer-</u>

\boxed{\boxed{L=\dfrac{g}{4\pi^2 f^2}}}

<u>Solution-</u>

The equation for time period of a simple pendulum is given by,

T=2\pi \sqrt{\dfrac{L}{g}}

Where,

T = Time period,

L = Length of the rod,

g = Acceleration due to gravity.

Frequency (f) of the pendulum is the reciprocal of its period, i.e

f=\dfrac{1}{T}\ \Rightarrow T=\dfrac{1}{f}

Putting the values,

\Rightarrow \dfrac{1}{f}=2\pi \sqrt{\dfrac{L}{g}}

\Rightarrow (\dfrac{1}{f})^2=(2\pi \sqrt{\dfrac{L}{g}})^2

\Rightarrow \dfrac{1}{f^2}=4\pi^2 \dfrac{L}{g}

\Rightarrow L=\dfrac{g}{4\pi^2 f^2}

8 0
3 years ago
Read 2 more answers
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