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astra-53 [7]
3 years ago
12

A CD store sold 3,467 CDs in 7 days. about the same number of CDs were sold each day. About how many CDs did the store sell each

day?
Mathematics
1 answer:
sergey [27]3 years ago
8 0
The store sold 495 CDs
You might be interested in
Differentiating Exponential Functions In Exercise, find the derivative of the function.<br> y = x2ex
I am Lyosha [343]

Answer:

y' = xe^{x}(2 + x)

Step-by-step explanation:

If we have a product function y, in the following format

y = f(x)*g(x)

This function has the following derivative

y' = f'(x)*g(x) + g'(x)*f(x)

In this problem, we have that:

y = x^{2}e^{x}

So

f(x) = x^{2}, f'(x) = 2x, g(x) = e^{x}, g'(x) = e^{x}

The derivative of the function is:

y' = f'(x)*g(x) + g'(x)*f(x)

y' = 2xe^{x} + x^{2}e^{x}

y' = e^{x}(2x + x^{2})

y' = xe^{x}(2 + x)

8 0
3 years ago
In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were o
ad-work [718]

The question is incomplete. Here is the complete qeustion.

In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were obtained: 0.10 0.13 0.16 0.15 0.14 0.008 0.15

(a) Construct a 99% confidence interval for the mean nitrogen-oxide emissions of all cars.

(b) If the EPA requires that nitrogen-oxide emissions be less than 0.165 g/mi, based on the 99% confidence interval in (a), can we safely conclude that this requirement is being met?

Answer: (a) 0.089 ≤ μ ≤ 0.171

(b) No

Step-by-step explanation:

(a) To determine the confidence interval, first calculate the mean (X) and standard deviation (s) of the sample

X = \frac{0.1+0.13+0.16+0.15+0.14+0.08+0.15}{7}

X = 0.13

s = \sqrt{\frac{(0.1-0.13)^{2} + (0.13 - 0.13)^{2} + ... + (0.15 - 0.13)^{2}}{7-1} }

s = 0.029

The degrees of freedom is

N - 1 = 7 - 1 = 6

And since the confidence is of 99%:

α = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005

The t-test statistics for t_{6,0.005} is 3.707

(Value found in the t-distribution table)

Now, calculate Error:

E = t_{6,0.005} . \frac{s}{\sqrt{N} }

E = 3.707. \frac{0.029}{\sqrt{7} }

E = 0.041

The interval will be:  

0.13 - 0.041 ≤ μ ≤ 0.13+0.041

0.089 ≤ μ ≤ 0.171

(b) No, because according to the interval, the nitrode-oxide emissions range from 0.089 to 0.171, which is greater than required by EPA.

7 0
4 years ago
Plz help me with this problem!!
elena-14-01-66 [18.8K]

there coul be an infinite amount of answers so anything above 73 should work as long AS ITS ABOVE 73

4 0
3 years ago
Cora chose to have her birthday party
kramer

$35, sorry if I’m wrong

6 0
3 years ago
What is 6 1/2 times 77
PSYCHO15rus [73]
500\frac{1}{2}
Hope this helps!
7 0
3 years ago
Read 2 more answers
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