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grandymaker [24]
3 years ago
15

Evaluate the expression ab for a = 10 and b = 4.

Mathematics
2 answers:
xz_007 [3.2K]3 years ago
7 0

A x B

= 10 x 4

= 40

Therefore answer is 40

PtichkaEL [24]3 years ago
4 0

a/b

= 10/4

=  5/2

Hope this helps... :)

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BRAINLIEST PLEASE HELP ASAP!!!!
Mashcka [7]

For this case we have that by definition of power properties it is fulfilled that:

(a ^ n) ^ m = a ^ {n * m}

We must rewrite the following function:

f (x) = 7 (\frac {1} {2})^{3x}}

Using the mentioned property we have:

f (x) = 7 ((\frac {1} {2}) ^ 3) ^ x

Solving the operation within the parenthesis we have:

f (x) = 7 (\frac {1 ^ 3} {2 ^ 3}) ^{ x}\\f (x) = 7 (\frac {1} {8})^{x}

Thus, the correct option is option B

ANswer:

Option B

5 0
3 years ago
-3 + 42 &gt; 3<br><br> x &gt; 15<br> x &lt; 15<br> x &gt; 13<br> x &lt; 13
Ilya [14]

Answer:

Where's the x?

I am assuming that the problem must be

-3 + 42 > 3x

if that is true, then

<u>x < 13</u>.

As this statement is, -3 + 42 > 3 is a true statement because -3 + 42 = 39, and

39 > 3 [39 is always greater than 3]

Step-by-step explanation:

Let me know if you need a step by step.

8 0
2 years ago
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assuming the growth rate stays constant at 1.2%, how many doublings would take place in a 500- year period?
Juliette [100K]

9514 1404 393

Answer:

  8.6

Step-by-step explanation:

The growth factor per year is 1.012, so in 500 years is ...

  1.012^500

The number of doublings is the solution to ...

  2^n = 1.012^500

Taking logarithms, we have ...

  n·log(2) = 500·log(1.012)

  n = 500·log(1.012)/log(2) ≈ 8.6046

About 8.6 doublings will take place in 500 years.

3 0
2 years ago
The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.10
Salsk061 [2.6K]

For the bell-shaped graph of the normal distribution of weights of Hershey kisses, the area under the curve is 1, the value of the median and mode both is 4.5338 G and the value of variance is 0.0108.

In the given question,

The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.1039 G.

We have to find the answer of many question we solve the question one by one.

From the question;

Mean(μ) = 4.5338 G

Standard Deviation(σ) = 0.1039 G

(a) We have to find for the bell-shaped graph of the normal distribution of weights of Hershey kisses what is the area under the curve.

As we know that when the mean is 0 and a standard deviation is 1 then it is known as normal distribution.

So area under the bell shaped curve will be

\int\limits^{\infty}_{-\infty} {f(x)} \, dx= 1

This shows that that the total area of under the curve.

(b) We have to find the median.

In the normal distribution mean, median both are same. So the value of median equal to the value of mean.

As we know that the value of mean is 4.5338 G.

So the value of median is also 4.5338 G.

(c) We have to find the mode.

In the normal distribution mean, mode both are same. So the value of mode equal to the value of mean.

As we know that the value of mean is 4.5338 G.

So the value of mode is also 4.5338 G.

(d) we have to find the value of variance.

The value of variance is equal to the square of standard deviation.

So Variance = (0.1039)^2

Variance = 0.0108

Hence, the value of variance is 0.0108.

To learn more about normally distribution link is here

brainly.com/question/15103234

#SPJ1

3 0
9 months ago
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