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olasank [31]
3 years ago
9

4. Gordon worked 621 hours in the last six months. He worked the same number

Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
4 0

Answer:

103.5 hours

Step-by-step explanation:

621/6 = 103.5 hours each month

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Bob sets the price of the plants according to their height in inches. A plant that is 9.5 inches high costs $2.28 in his store.
alekssr [168]
2.28 devide 9.5
=0.24
7 0
3 years ago
What is the value of the expression f(-3) + 2f(-1) - f (4)<br><br> PLEASE HELP NO LINKS
Bas_tet [7]

Answer:

22

Step-by-step explanation:

Find f(-3) :

f(x) = x² + 2x for x ≤ -3

That means for those values of x, less than or equal to -3, this is the function.

So, f(-3) = (-3)² + 2(-3) = 9 - 6 = 3

Now, f(-1) :

From the given data, we see it is: f(x) = 2 (\frac{1}{3} )^{2x}

We take this because -1 lies between -3 and 4.

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

 => f = (-1) = 2(3)^{2} = 2 (9) = 18

For f(4) :

Clearly, the function is: f (x) = \frac{2x - 5}{x-7}

f (4) = \frac{2(4)-5}{4-7} = \frac{3}{-3} = -1

Therefore, <u>f(-3) + f(-1) - f(4) = 3 + 18 - (-1) = 3 + 18 + 1 = 22.</u>

5 0
3 years ago
Plsssssss help i need it.
riadik2000 [5.3K]
I would say decreasing!
7 0
3 years ago
An unprepared student makes random guesses for the ten​ true-false questions on a quiz. Find the probability that there is at le
hoa [83]

Answer:

0.999

Step-by-step explanation:

At least 1 correct means, 1 correct, 2 correct, 3 correct ... until 10 correct. That would be a long process to calculate.

<em>Instead we use the complement rule to calculate.</em>

<em>P(x\geq1)=1-P(x</em>

<em />

<em>So we need to find P(x<1). So this is getting 0 answers correct, or 10 incorrect.</em>

<em />

In true false question, probablity of correct is 1/2 and incorrect is 1/2, hence,

Probability of 10 incorrect is (1/2)^10

Thus,

P(x\geq1)=1-(\frac{1}{2})^{10}=0.999

So the answer is 0.999 (rounded to nearest thousandth)

3 0
3 years ago
The coefficient of -x is 1
juin [17]

Answer:

it -1

Step-by-step explanation:

3 0
3 years ago
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