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Snowcat [4.5K]
2 years ago
7

Kate gave 5/6 of her CD's to Sue

Mathematics
1 answer:
Alexus [3.1K]2 years ago
7 0
Y=0.72(3) i think tht is the answer
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How many different ways can you write a fraction that has a numerator of 2 as a sum of fractions
Rzqust [24]
At least you can write it as the sum of half of it twice, that is:
2/x = 1/x + 1/x
whatever x number you have, for example:
2/5 = 1/5 + 1/5

But then, you can express a fraction in many different ways, for example again:
2/5 = <span>1/5 + 1/5 = 2/10 + 2/10
</span>= 1/10 + 1/10 + 1/10 + 1/10
so there are an infinite ways of expressing such a fraction.
7 0
3 years ago
Graph the equation F(x)=(1/3)^x
scoray [572]

The value of a=9,b=3 and c=1

Step-by-step explanation:

The graph shown is for function f(x)

The given function is f(x)=(\frac{1}{3}) ^{x}

Table says,

X      Y

-2      a

1        b

0       c

1     (1/3)

2    (1/9)

To find value of a:

From table, for output value a, input value, x=(-2)

we can write f(-2)=a

Therefore,

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

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

a=(\frac{1}{3^{(-2)}})

a=3^{(2)}

a=9

To find value of b:

From table, for output value a, input value, x=(-1)

we can write f(-1)=b

Therefore,

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

f(-1)=(\frac{1}{3}) ^{(-1)}

b=(\frac{1}{3^{(-1)}})

b=3^{(1)}

b=3

To find value of c:

From table, for output value a, input value, x=(0)

we can write f(0)=c

Therefore,

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

f(0)=(\frac{1}{3}) ^{(0)}

c=(\frac{1}{3^{(0)}})

c=1

5 0
2 years ago
(7x10*5)x(3x10*2) in standerd form
marishachu [46]

Answer:

21,000

Step-by-step explanation:

This is more of straightforward multiplication, so I cannot explain much! If you want me to try to do more, then just comment.

(7 * 10 * 5) * (3 * 10 * 2)

= (70 * 5) * (30 * 2)

= (350) * (60)

= 21,000

Thusly, in standard form, the equation given is \boxed{21,000}.

Hope this helps! (:

7 0
3 years ago
The average performance rating of the employees at Company A is 6.5 on a scale from 1 to 10 (10 being highest). The standard dev
djyliett [7]

Answer:

The best choice would be hiring a random employee from company A

Step-by-step explanation:

<em>Supposing that the performance rating of employees follow approximately a normal distribution on both companies</em>, we are interested in finding what percentage of employees of each company have a performance rating greater than 5.5 (which is the mean of the scale), when we measure them in terms of z-scores.

In order to do that we standardize the scores of both companies with respect to the mean 5.5 of ratings

The z-value corresponding to company A is

z=\frac{\bar x-\mu}{s}

where

\bar x = mean of company A

\mu = 5.5 (average of rating between 1 and 10)

s = standard deviation of company A

z=\frac{\bar x-\mu}{s}=\frac{6.5-5.5}{2.1}=0.7142

We do the same for company C

z=\frac{\bar x-\mu}{s}=\frac{7.4-5.5}{6.8}=0.2794

This means that 27.49% of employees of company C have a performance rating > 5.5, whereas 71.42% of employees of company B have a  performance rating > 5.5.

So, the best choice would be hiring a random employee from company A

4 0
3 years ago
Perform the modular arithmetic. (7 · 15) (mod 4)
Reika [66]

We could do it by first writing

7\cdot15=(4+3)\cdot(4+11)=4^2+14\cdot4+33

so that

7\cdot15\equiv4^2+14\cdot4+33\equiv33\pmod4

and since 33=32+1=8\cdot4+1, we have

7\cdot15\equiv33\equiv1\pmod4

# # #

Alternatively, we can just compute the product and take it mod 4:

7\cdot15=105=104+1=26\cdot4+1

and so we get the same result of 1.

4 0
3 years ago
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