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Natalka [10]
3 years ago
12

The radius of a circle is 14 cm. Determine the area of the circle.

Mathematics
1 answer:
iris [78.8K]3 years ago
3 0
Area of a Circle = πr²
A = 3.14 * (14)²
A = 3.14 * 196
A = 615.44 cm²

In short, Your Answer would be: Option A

Hope this helps!
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What is the answer to 13 in precise detail?
lyudmila [28]

Well, I don't know.  Let's see . . . . .

First draw: One 'M' available out of 11 letters.   Probability of picking it = 1/11.
2nd draw:  Four 'I's available out of 10 letters.  Probability of picking one = 4/10.
3rd draw:  Four 'S's available out of 9 letters.   Probability of picking one = 4/9.
4th draw:  Three 'S's available out of 8 letters. Probability of picking one = 3/8.
5th draw:  Three 'I's available out of 7 letters.  Probability of picking one = 3/7.
6th draw:  Two 'S's available out of 6 letters.   Probability of picking one = 2/6.
7th draw:  One 'S' available out of 5 letters.     Probability of picking it = 1/5.
8th draw:  Two 'I's available out of 4 letters.     Probability of picking it = 2/4.
9th draw:  Two 'P's available out of 3 letters.   Probability of picking one = 2/3.
10th draw: One 'P' available out of 2 letters.   Probability of picking it = 1/2.
11th draw: One letter left.  It is an 'I'.  Probability of picking it = 1 .

Probability of all of those draws in order =

    (1/11) x (4/10) x (4/9) x (3/8) x (3/7) x (2/6) x (1/5) x (2/4) x (2/3) x (1/2) x (1) =

           1,152 / 39,916,800 =

                 1 / 34,650 =

                                     0.00002886 =

                                   <em>  0.002886 percent</em> (rounded)

Not a good bet.  (But better than the lottery.)


7 0
4 years ago
The sales of a grocery store had an average of $7,000 per day. The store introduced several advertising campaigns in order to in
Sati [7]

Answer:

a.The advertising campaigns increased the average daily sales.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the advertising campaigns increased the average daily sales.

Then, the null and alternative hypothesis are:

H_0: \mu=7000\\\\H_a:\mu> 7000

The significance level is 0.05.

The sample has a size n=100.

The sample mean is M=7280.

The standard deviation of the population is known and has a value of σ=1000.

We can calculate the standard error as:

\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{1000}{\sqrt{100}}=100

Then, we can calculate the z-statistic as:

z=\dfrac{M-\mu}{\sigma_M}=\dfrac{7280-7000}{100}=\dfrac{280}{100}=2.8

This test is a right-tailed test, so the P-value for this test is calculated as:

\text{P-value}=P(z>2.8)=0.0026

As the P-value (0.0026) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the advertising campaigns increased the average daily sales.

7 0
3 years ago
Round 0.007492 to four decimal places.
abruzzese [7]
Answer: 0.0075

Just round up the 4 to 5 then u got your answer.
8 0
2 years ago
List all 6 whole numbers between -3.5 and 5.5
Alenkinab [10]

Answer012345

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Anyone know any equivelant expressions to 2(4x−3)+3x−1?
Westkost [7]
Sure, each of the following lines.
8x-6+3x-1
11x-6-1
and, 11x-7
8 0
3 years ago
Read 2 more answers
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