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Anestetic [448]
4 years ago
5

3. After completing a statistical analysis of a survey of 40 students, the principal of North High School made the following con

clusion: reject the null hypothesis; there is convincing evidence that more than 50% of students support a schedule change to have lunch occur earlier in the day. Which error could have been committed? (A) Type I error: Conclude that more than 50% of students want earlier lunch, when 50% or less want earlier lunch. (B) Type I error: Conclude that more than 50% of students want earlier lunch, when more than 50% want earlier lunch. (C) Type II error: Fail to reject that 50% of students want earlier lunch, when more than 50% want earlier lunch. (D) Type II error: Fail to reject that 50% of students want earlier lunch, when 50% or less want earlier lunch. (E) Type II error: Fail to reject that more than 50% of students want earlier lunch, when 50% or less want earlier lunch.
Mathematics
1 answer:
iragen [17]4 years ago
4 0

Answer: A

Step-by-step explanation:

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What times what equals 234, but then if you add them you get 31
USPshnik [31]
13*18
x*y =234
x+y=31

we know the 2 # have to be double digits and both between 10-20

find 2 values that * to a number that ends in 4. Those are the ones value

1*4   11 and 14
2*2   12 and 12
3*8   13 and 18

only 13 and 18 add up to 31 and multiply to 234



6 0
4 years ago
(HELP ME 15 PTS)
kap26 [50]

Answer:   \bold{\dfrac{27x^2y^3}{16}}

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

\dfrac{(3x^2y^5)^3}{(2xy^3)^4}\\\\\\=\dfrac{3^3\cdot x^{2\cdot 3}\cdot y^{5\cdot 3}}{2^4\cdot x^4\cdot y^{3\cdot 4}}\\\\\\=\dfrac{27\cdot x^6\cdot y^{15}}{16\cdot x^4\cdot y^{12}}\\\\\\=\dfrac{27\cdot x^{6-4}\cdot y^{15-12}}{16}\\\\\\=\dfrac{27\cdot x^2\cdot y^3}{16}

8 0
3 years ago
Helpppppppppppppppppp
Anuta_ua [19.1K]

Option A:

\left(x^{4}\right)\left(3 x^{3}-2\right)\left(4 x^{2}+5 x\right)=12 x^{9} +15 x^{8} -  8 x^{6}-10 x^{5}

Solution:

Given expression \left(x^{4}\right)\left(3 x^{3}-2\right)\left(4 x^{2}+5 x\right).

To find the product of the above expression:

\left(x^{4}\right)\left(3 x^{3}-2\right)\left(4 x^{2}+5 x\right)

First multiply first two factors with each term.

           =(x^{4} \times 3 x^{3}- x^{4} \times 2 ) \left(4 x^{2}+5 x\right)

Using exponent rule: a^m \cdot a^n=a^{m+n}

            =(3 x^{7}- 2x^{4} ) \left(4 x^{2}+5 x\right)

Now multiply these two factors with each term.

            =3 x^{7} (4 x^{2}+5 x)- 2x^{4} \left(4 x^{2}+5 x\right)

            =(4 x^{2} \times 3 x^{7} +5 x \times 3 x^{7} )-  \left(4 x^{2} \times 2x^{4}+5 x \times 2x^{4}\right)

Using exponent rule: a^m \cdot a^n=a^{m+n}

            =(12 x^{9} +15 x^{8} )-  (8 x^{6}+10 x^{5})

           =12 x^{9} +15 x^{8} -  8 x^{6}-10 x^{5}

\left(x^{4}\right)\left(3 x^{3}-2\right)\left(4 x^{2}+5 x\right)=12 x^{9} +15 x^{8} -  8 x^{6}-10 x^{5}

Hence option A is the correct answer.

4 0
4 years ago
Describe the relationship between the values of the digits in the number 9999999
Gemiola [76]

Answer:

  • <u>Each digit has a value ten times the value of the digit to its right.</u>

Explanation:

The given number is 9,999,999.

Each digit has a value according to its place.

The right most 9 is in the place of the ones and its value is 9 × 10⁰ = 9 × 1 = 9.

The second 9 from the right is in the place of the tens and its value is 9 × 10 = 90.

The third 9 from the right end is in the place of the hundreds and its value is 9 × 100 = 900.

The next 9 is in the place of the thousands, so its value is 9,000.

So, each 9 has a value ten times the 9 to its right.

6 0
3 years ago
What are the roots of the equation: X^2 - 10x -20 =20
Alexxx [7]
A = 1  b = -10  c = -20
Using the quadratic formula,
x = [--10 +-sqr root (100 -4*1*-20)] / 2*1
x = [10 +- sqr root (180)] / 2
x = 10/2 +- (sqr root (180) / 2)
x = 5 +- (3 * sqr root (20) / 2)
x = 5 +- (6 * sqr root (5) / 2)
x = 5 +- 3 * sqr root (5)

So the answer is #3

8 0
3 years ago
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