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2 years ago

Mia said that 2 - 3a = 11 and 3a -2 +1 have the same solution. Is she correct? Explain.

Mathematics

1 answer:

Sedaia [141]2 years ago

5 0

2-3a=11
Minus 2 from both sides
-3a=9
a=-3
3a-3=1
Plus 3 to both sides
3a=4
a=4/3
No

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For f(x) = 4x+1 and g(x) = x^2 -5, find (f-g)(x)

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Answer:

(f-g)(x) is given by -x^2+4x+6

Step-by-step explanation:

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1 year ago

Evaluate the expression 2(8 - 4)^2 - 10 / 2

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John and Kamira are playing a game. John's score (J) and Kamira's score (K) after round 1 are shown on the number line.

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Complete question :

John and Kamira are playing a game. John's score (J) and Kamira's score (K) after round 1 are shown on the number line.

The score recorded at the end of the first round is 2.

What could this score represent?

Options :

the sum of John's score and Kamira's score

the difference between John's score and Kamira's score

the absolute value of the difference of John's score and Kamira's score

the sum of the absolute value of John's score and the absolute value of Kamira's score

Answer:

the sum of John's score and Kamira's score

Step-by-step explanation:

Given that :

The score recorded at the end of the round is 2

From the attached number line :

John's score = - 5

Kamira's score = 7

Sum = - 5 + 7 = 2

Difference = - 5 - 7 = - 12

Sum of absolute value = 5 + 7 = 12

absolute value of difference = 12

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2 years ago

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Which Function Has the greatest rate of change?

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2 years ago

There were 800 math instructors at a mathematics convention. Forty instructors were randomly selected and given an IQ test. The

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Answer:

95% confidence interval for the mean of the 800 instructors is [126.80 , 133.20].

Step-by-step explanation:

We are given that there were 800 math instructors at a mathematics convention.

Forty instructors were randomly selected and given an IQ test. The scores produced a mean of 130 with a standard deviation of 10.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean score = 130

s = sample standard deviation = 10

n = sample of instructors = 40

$\mu$ = population mean of 800 instructors

Here for constructing 95% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.0225 < $t_3_9$ < 2.0225) = 0.95 {As the critical value of t at 39 degree of

freedom are -2.0225 & 2.0225 with P = 2.5%}

P(-2.0225 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.0225) = 0.95

P( $-2.0225 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.0225 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.0225 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.0225 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.0225 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.0225 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $130-2.0225 \times {\frac{10}{\sqrt{40} } }$ , $130+2.0225 \times {\frac{10}{\sqrt{40} } }$ ]

= [126.80 , 133.20]

Therefore, 95% confidence interval for the mean of the 800 instructors is [126.80 , 133.20].

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