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3 years ago

Passes through (2,2), slope = 3

Mathematics

1 answer:

nikdorinn [45]3 years ago

5 0

Answer:

y - 2 = 3(x - 2)

Step-by-step explanation:

Hello!

Please consider using the point-slope formula for the equation of a straight line, because in this case it's the easiest way to go.

(2,2) is one point on the line, and the slope is m = -3.

Thus, y - k = m(x - h) becomes y - 2 = 3(x - 2).

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A painter is placing a ladder to reach the third story window, which is 34 feet above the ground and

denis-greek [22]

Answer:

34 times 70

Step-by-step explanation:

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3 years ago

A cereal box is 8 in by 2in by 12in what is the surface area.

LUCKY_DIMON [66]

192 in2

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4 0

2 years ago

PLEASE HELP ASAP

Elena L [17]

36.08 = 8.8t

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3 years ago

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

Then, if $|x-5|$ , then $|f(x)-14|$ . So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$ . The maximum value of delta is $\frac{\varepsilon}{3}$ .

By definition, this procedure proves that $\lim_{x\to 5}f(x) = 14$ . Note that f(5)=14, so this proves that f is continous at x=5.

3 0

2 years ago

A student wants to estimate the mean score of all college students for a particular exam. First use the range rule of thumb to m

Wewaii [24]

Answer:

See explanation below

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

range = 2200 - 300 = 1900

To find standard deviation, we have:

standard deviation = range/4 = $\frac{1900}{4}$ = 475

The range rule of thumb estimate for the standard deviation is 475

Given:

Standard deviation, $\sigma$ = 475

Margin of Error, ME = 100

$\alpha$ = 1 - 0.90 = 0.10

Za/2 = Z0.05 = 1.64

Find sample size, n:

n ≥ $[Z_\alpha_/_2 * (\frac{\sigma}{ME})]^2$

n ≥ $[1.64 * (\frac{475}{100})]^2$

n ≥ 60.68

≈ 61

Minimun sample size,n = 61

5 0

3 years ago