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3 years ago

What is the slope of (1,5) and (4,-2)

Mathematics

1 answer:

andriy [413]3 years ago

6 0

Answer:-2-5÷4-1=-1

Step-by-step explanation:

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a train leaves a station and travels north at a speed of 105mph. two hours later, a second train leaves on a parallel track and

meriva

Let the second train catch up after x hours

Distance = speed X time 

135x = 105 ( 2 + x) 

135x = 210 + 105x 

30x = 210 

x = 7 hours (after 2nd train has started ) 
ANSWER 7 X 135 = 945 miles from the starting point 

CHECK 
7 X 135 = 945 miles 

9 X 105 = 945 miles

8 0

3 years ago

How do you change 0.81 into a fraction?

fredd [130]

Make use of your knowledge of place value.

eighty-one hundredths = 81/100

Numerator and denominator have no common factors, so this fraction cannot be reduced.

5 0

3 years ago

Write the equation of the absolute value function y = 3|x| translated right 8 units and up 1 unit.

Reil [10]

Answer:

Step-by-step explanation:

y=3|x|, shift right 8 units

y=3|x-8|, shift up 1 unit

y=3|x-8|+1

6 0

2 years ago

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

Leya [2.2K]

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].

5 0

3 years ago

Read 2 more answers

12-(4y+8)=0.5(8y-16)

m_a_m_a [10]

Answer:

3/2

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers