Ask question

Ask question

All categories

3 years ago

12

A camera has a listed price of $876.99 before tax. If the sales tax rate is 6.75%, find the total cost of the camera with sales

tax included.
Round your answer to the nearest cerit, as necessary

1 answer:

avanturin [10]3 years ago

5 0

Answer:

$936.18

Byehave fun k bye

You might be interested in

Find the volume of this right rectangular prism. [Type your answer as a number.]

irakobra [83]

There is no picture

3 0

4 years ago

Simplify (a^2b^3)^5 a^10b^15 a^7b^8 (ab)^30 a^10b^8

butalik [34]

The answer to your problem is a^67b^76

5 0

3 years ago

What's 2+2 I really need this answer plz help ASAP

kondor19780726 [428]

Answer:

4

Step-by-step explanation:

2+2=4

6 0

3 years ago

Read 2 more answers

A survey conducted by the Consumer Reports National Research Center reported, among other things, that women spend an average of

Nookie1986 [14]

Answer:

(a) The probability that a randomly selected woman shop exactly two hours online is 0.217.

(b) The probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c) The probability that a randomly selected woman shop less than 5 hours online is 0.9922.

Step-by-step explanation:

Let <em>X</em> = time spent per week shopping online.

It is provided that the random variable <em>X</em> follows a Poisson distribution.

The probability function of a Poisson distribution is:

$P (X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0,1,2,...$

The average time spent per week shopping online is, <em>λ </em>= 1.2.

(a)

Compute the probability that a randomly selected woman shop exactly two hours online over a one-week period as follows:

$P (X=2)=\frac{e^{1.2}(1.2)^{2}}{2!} =0.21686\approx0.217$

Thus, the probability that a randomly selected woman shop exactly two hours online is 0.217.

(b)

Compute the probability that a randomly selected woman shop 4 or more hours online over a one-week period as follows:

P (X ≥ 4) = 1 - P (X < 4)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{1.2}(1.2)^{0}}{0!}-\frac{e^{1.2}(1.2)^{1}}{1!}-\frac{e^{1.2}(1.2)^{2}}{2!}-\frac{e^{1.2}(1.2)^{2}}{3!}\\=1-0.3012-0.3614-0.2169-0.0867\\=0.0338$

Thus, the probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c)

Compute the probability that a randomly selected woman shop less than 5 hours online over a one-week period as follows:

P (X < 5) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

$=\frac{e^{1.2}(1.2)^{0}}{0!}+\frac{e^{1.2}(1.2)^{1}}{1!}+\frac{e^{1.2}(1.2)^{2}}{2!}+\frac{e^{1.2}(1.2)^{3}}{3!}+\frac{e^{1.2}(1.2)^{4}}{4!}\\=0.3012+0.3614+0.2169+0.0867+0.0260\\=0.9922$

Thus, the probability that a randomly selected woman shop less than 5 hours online is 0.9922.

8 0

4 years ago

What ratio is equivalent to 1:4

steposvetlana [31]

Answer:

Fractions equivalent to 1/4 are 2/8, 3/12, 4/16, 5/20,

Step-by-step explanation:

I searched it up : )

8 0

3 years ago