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3 years ago

Write the equation -3x+2y=7 in slope intercept form.

Mathematics

2 answers:

Sladkaya [172]3 years ago

8 0

Answer:

$y=\frac{3}{2}x+\frac{7}{2}$

Step-by-step explanation:

Slope-intercept form is given by y = mx + b

We need to re-arrange the equation given to have y to the left side of the equal sign (in other words, solve for y). Steps are shown below:

$-3x+2y=7\\2y=3x+7\\y=\frac{3x+7}{2}\\y=\frac{3x}{2}+\frac{7}{2}\\y=\frac{3}{2}x+\frac{7}{2}$

the last answer is correct.

Mrrafil [7]3 years ago

4 0

For this case we have that the line equation of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut point with the y axis.

We have the equation:

$-3x + 2y = 7$

By adding 3x to both sides of the equation we have:

$-3x + 3x + 2y = 7 + 3x\\2y = 3x + 7$

Dividing between 2 on both sides of the equation:

$\frac {2y} {2} = \frac {3x} {2} + \frac {7} {2}\\y = \frac {3x} {2} + \frac {7} {2}$

ANswer:

Option D

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3 years ago

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What is the value of r to the nearest tenth?

Tatiana [17]

Answer:

Hence r=10.3 cms (rounded to tenth place)

Step-by-step explanation:

In Triangle KLN ,

KN = 5 Cms

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As the Perpendicular dropped from the center to the chord bisects it.

LM = 18 cm

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We apply Hypotenuse Formula in this

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$c=\sqrt{a^2+b^2}$

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a=5

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3 years ago

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3 years ago

A researcher wishes to estimate the average blood alcohol concentration (BAC) for drivers involved in fatal accidents who are f

Mnenie [13.5K]

Answer:

A 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

Step-by-step explanation:

We are given that a researcher randomly selects records from 60 such drivers in 2009 and determines the sample mean BAC to be 0.16 g/dL with a standard deviation of 0.080 g/dL.

Firstly, the pivotal quantity for finding the 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 BAC = 0.16 g/dL

s = sample standard deviation = 0.080 g/dL

n = sample of drivers = 60

$\mu$ = population mean BAC in fatal crashes

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, a 90% confidence interval for the population mean, $\mu$ is;

P(-1.672 < $t_5_9$ < 1.672) = 0.90 {As the critical value of t at 59 degrees of

freedom are -1.672 & 1.672 with P = 5%} P(-1.672 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.672) = 0.90

P( $-1.672 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $0.16-1.672 \times {\frac{0.08}{\sqrt{60} } }$ , $0.16+1.672 \times {\frac{0.08}{\sqrt{60} } }$ ]

= [0.143, 0.177]

Therefore, a 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

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3 years ago

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suter [353]

Answer:

-7

Step-by-step explanation:

-21.00/3.00

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2 years ago