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deff fn [24]
3 years ago
6

Which of the following samples would meet the "random" condition?

Mathematics
1 answer:
Troyanec [42]3 years ago
7 0

Answer:

Brother there's nothing there

Step-by-step explanation:

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ankoles [38]
Can you answer the question in blue please
6 0
3 years ago
. Denise has multiplied (2r – 3)(2r + 3) to get 4r2 – 9. After looking at her work, Hyder says her solution must be incorrect be
makvit [3.9K]

Answer:

The answer is 4r² – 9

Denise answer is correct.

Step-by-step explanation:

One of the methods of solving the given equation [(2r – 3)(2r + 3)] is the FOIL method

Using the FOIL method (Used for multiplying binomials)

F stands for first

O stands for outer

I stands for inner

L stands for last

In this method, it means we will first multiply the first terms followed by multiplying the outer term, then multiplying the inner terms and finally, the last terms.

So, for the equation  (2r – 3)(2r + 3)

First : (2r)(2r) = 4r²

Outer : (2r)(+3) = 6r

Inner : (–3)(2r) =  – 6r

Last :  (–3)(+3) =  – 9

Adding the results we get

4r² + 6r – 6r – 9 (+6r and – 6r cancels out)

= 4r² – 9

∴ (2r – 3)(2r + 3) = 4r² – 9

Hence, Denise answer is correct

5 0
3 years ago
Researchers have claimed that the average number of headaches per student during a semester of Statistics is 11. In a sample of
SSSSS [86.1K]

Answer:

A) H0: μ = 11 vs. Ha: μ > 11

Step-by-step explanation:

The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.

Therefore, for the case above;

The null hypothesis is that the average number of headaches per student during a semester of Statistics is 11.

H0: μ = 11

The alternative hypothesis is that the average number of headaches per student during a semester of Statistics is greater than 11.

Ha: μ > 11

4 0
4 years ago
Write the quadratic equation in general form.
MrRa [10]

Answer: C

<u>Step-by-step explanation:</u>

(x - 3)² = 0

(x - 3)(x - 3) = 0

x(x - 3)  -3(x - 3) = 0

x² - 3x  -3x + 9  = 0

x² - 6x + 9  = 0

4 0
3 years ago
The area of a rectangular field is 1000 yd2.Two parallel sides are fenced with aluminum at $15/yd. One of the remaining sides is
ioda
Coolio

lw=1000
lets assume that the aluminum sides are the legnth
2 sides
2*15=30
30l=cost of aluminum

steel=10w
wood=(w-10)5=5w-50
we can eliminate the w by solving for w in first relation

lw=1000
divide both sides by l
w=1000/l
sub that for w

10(1000/l)=steel
5(1000/l)-50=wood

wood cost+steel+aluminum cost=total cost
10(1000/l)+5(1000/l)-50+30l=1525
(10000/l)+(5000/l)-50+30l=1525
(15000/l)-50+30l=1525
add 50 to both sides
(15000/l)+30l=1575
times both sides by l
15000+30l²=1575l
minus 1575l from both sides
30l²-1575l+15000=0
factor
(15)(l-40)(2l-25)=0
set each equal to 0
l-40=0
l=40

2l-25=0
2l=25
l=12.5



the there are 2 possible combinations

aluminum=40yd and steel=25yd or
aluminum=12.5yd and steel=80ft
both yield 1000yd² and cost $1525
5 0
3 years ago
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