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Paul [167]
3 years ago
9

Question 11

Mathematics
1 answer:
shepuryov [24]3 years ago
3 0

Answer:

A. x^27/10

B. 1/x^-27/10

Step-by-step explanation:

Mathematically, if we have the same base, we simply add up the index

Thus, we have that;

x^1/5•x^5/2 = x^(1/5 + 5/2) = x^(2 + 25)/10 = x^27/10

A) In the first form, we have that;

x^27/10

a = 27/10

B) We simply negate what we have above in terms of the index according to the indices law below;

a^b = 1/a^-b

Hence, we have that;

1/x^-27/10

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How many of each color cubes are? .. answer please
ss7ja [257]

Answer:

2 cubes are red, 6 are blue, the rest are yellow so 8

Step-by-step explanation:

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3 years ago
Bo invests money in an account paying a simple interest of 8% per year. If no money will be added or removed from the investment
inn [45]

Answer:

Calculation:

First, converting R percent to r a decimal

r = R/100 = 8%/100 = 0.08 per year,

then, solving our equation

I = Prt (Where p=principle,r=rate,t=time)

I = p*0.08*1=0.08p

Step-by-step explanation:

3 0
3 years ago
Does anyone know this? I’m really stuck
S_A_V [24]

<em>p = 21</em>

Step-by-step explanation:

When you have a midsegment, the base is going to be two times as large as the midsegment. You can find the midsegment if you multiply it by two and set it equal to the base. I have attached a photo below that may help you.

This is the equation we will be using.

2(p-13)=p-5

Now let's solve! Let's start by distributing the 2 by what's in the parentheses.

2p-26=p-5

Subtract p from both sides.

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Add 26 to both sides.

<u>p = 21</u>

7 0
3 years ago
Read 2 more answers
One year had the lowest ERA​ (earned-run average, mean number of runs yielded per nine innings​ pitched) of any male pitcher at
lord [1]

Answer:

Thomas had the better year relative to their​ peers.

Step-by-step explanation:

<u>The complete question is</u>: One year Thomas had the lowest ERA​ (earned-run average, mean number of runs yielded per nine innings​ pitched) of any male pitcher at his​ school, with an ERA of 3.31. ​Also, Karla had the lowest ERA of any female pitcher at the school with an ERA of 3.02. For the​ males, the mean ERA was 4.837 and the standard deviation was 0.541. For the​ females, the mean ERA was 4.533 and the standard deviation was 0.539. Find their respective​ z-scores. Which player had the better year relative to their​ peers, or ​? ​(Note: In​ general, the lower the​ ERA, the better the​ pitcher.)

We are given that for the​ males, the mean ERA was 4.837 and the standard deviation was 0.541. For the​ females, the mean ERA was 4.533 and the standard deviation was 0.539.

As, we know that the z-score is calculated by the following formula;

                                 Z  =  \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = population mean

           \sigma = standard deviation

Now, firstly we will calculate the z score for Thomas;

z-score = \frac{X-\mu}{\sigma}

            = \frac{3.31-4.837}{0.541}  = -2.823

{Here, the mean ERA for the males was 4.837 and the standard deviation was 0.541}

Similarly, we will calculate the z score for Karla;

z-score = \frac{X-\mu}{\sigma}

            = \frac{3.02-4.533}{0.539}  = -2.807

{Here, the mean ERA for the females was 4.533 and the standard deviation was 0.539}

Now, it is stated in the question that the lower the​ ERA, the better the​ pitcher.

So, we can clearly see that Thomas had a lower ERA of z-score as -2.823 < -2.807. This means that Thomas had the better year relative to their​ peers.

5 0
3 years ago
a rectangular section of land has a length of 5•10^3 meters and a width of 6•10^2 meters. what is the area of land in square met
monitta
A = lw
A = (5 × 10^3)(6 × 10^2)
= 5000 × 600
= 300000
= 3 × 10^6 m^2
4 0
3 years ago
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