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nadya68 [22]
3 years ago
15

CAN SOMEONE PLEASE HELP ME WRITE A EQUATION FOR THIS SCENARIO, ILL GIVE BRAINLIST

Mathematics
2 answers:
OlgaM077 [116]3 years ago
7 0

Answer:

y = -5x + 65 or f(x) = -5x + 65

Step-by-step explanation:

The graph represents a linear function which means the function can be represented using y = xm + a, where <em>y</em> is the target, <em>x</em> is the domain, <em>m</em> is the slope, and <em>a</em> is the y-intercept.

According to the graph, the y-intercept is $65 which means a = 65

y = xm + 65

Since the x intercept is 13 that means when x = 13, then y = 0, plug those values in and solve for m to get the slope.

0 = 13m + 65

-65 = 13m

-65/13 = m

Which can be simplified to

-5 = m

Therefore the function for the graph is y = -5x + 65

patriot [66]3 years ago
5 0

Answer:

day 13

Step-by-step explanation:

the y axis represents the money. ans the x represents the days

so go down to 0 on the y acis and see what number the line matches with 0 on

in this case it matches with 13

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Sergeu [11.5K]
Your answer should be 0.3000
5 0
3 years ago
Read 2 more answers
WHAT IS THE ANSWER FOR 12 1/3 + 8 3/4 +17 2/8 +23 2/3=
gulaghasi [49]
12  \dfrac{1}{3}  + 8  \dfrac{3}{4}  +17  \dfrac{2}{8}  +23  \dfrac{2}{3}

Simplify
= 12  \dfrac{1}{3}  + 8  \dfrac{3}{4}  +17  \dfrac{1}{4}  +23  \dfrac{2}{3}

Change the denominators to be the same
= 12  \dfrac{1 \times 4}{3 \times 4}  + 8  \dfrac{3 \times 3}{4 \times 3}  +17  \dfrac{1 \times 3 }{4 \times 3}  +23  \dfrac{2 \times 4}{3 \times 4}

= 12  \dfrac{4}{12}  + 8  \dfrac{9}{12}  +17  \dfrac{3 }{12}  +23  \dfrac{8}{12}

Combine into single fraction
= 60  \dfrac{24}{12}

= 62

6 0
3 years ago
An automobile manufacturer is considering using robots for part of its assembly process. Converting to robots is an expensive pr
LenKa [72]

Answer:

(a) The correct option is: <em>H₀</em>: <em>p</em> = 0.02 vs. <em>Hₐ</em>: <em>p</em> < 0.02.

(b) Explained below.

(c) The better value of <em>α</em> will be 0.10.

Step-by-step explanation:

An automobile manufacturer is considering using robots for part of its assembly process only if there is strong evidence that the proportion of defective installations is less for the robots than for human assemblers.

To test whether the proportion of defective installations is less for the robots than for human assemblers use a single-proportion <em>z</em>-test.

(a)

The hypothesis can be defined as:

<em>H₀</em>: The proportion of defective installations is same for both the robots and  human assemblers, i.e. <em>p</em> = 0.02.

<em>Hₐ</em>: The proportion of defective installations is less for the robots than for human assemblers, i.e. <em>p</em> < 0.02.

The alternate hypothesis is the claim or the statement that is being tested.

In this case we need to test whether the proportion of defective installations is less for the robots than for human assemblers or not, so that the manufacturer can decide whether they want to apply the conversion.

Thus, the correct option is:

<em>H₀</em>: <em>p</em> = 0.02 vs. <em>Hₐ</em>: <em>p</em> < 0.02.

(b)

A type I error occurs when we discard a true null hypothesis and a type II error is made when we fail to discard a false null hypothesis.

In this case a type I error will be committed if conclude that the proportion of defective installations is less for the robots than for human assemblers when in fact it is not.

And a type II error will be committed if we fail to conclude that proportion of defective installations is less for the robots than for human assemblers.

(c)

The power of the test is the probability of rejecting a false null hypothesis.

The power of the test sis affected by the significance level of the test (<em>α</em>).

Lesser the significance level of the test the lesser is the power of the test.

If the value of <em>α</em> is reduced from 0.05 to 0.01 then the region of acceptance will increase. This implies that there is low probability of rejecting the null hypothesis even when it is false.

So higher the value of <em>α</em> the higher is the probability of making a correct decision.

Thus, the better value of <em>α</em> will be 0.10.

3 0
3 years ago
What is the number of possible combinations of 14 objects taken 3 at a time?
maria [59]

Answer:

364.

Step-by-step explanation:

14C3 = 14! / 11! 3! which simplifies to:

14*13*12 /3*2*1

= 14*13* 2

= 364.

6 0
3 years ago
Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.
Leni [432]

Answer:

a)

g(2.9) \approx -6.6

g(3.1) \approx -3.4

b)

The values are too small since g'' is positive for both values of x in. I'm speaking of the x values, 2.9 and 3.1.

Step-by-step explanation:

a)

The point-slope of a line is:

y-y_1=m(x-x_1)

where m is the slope and (x_1,y_1) is a point on that line.

We want to find the equation of the tangent line of the curve g at the point (3,-5) on g.

So we know (x_1,y_1)=(3,-5).

To find m, we must calculate the derivative of g at x=3:

m=g'(3)=(3)^2+7=9+7=16.

So the equation of the tangent line to curve g at (3,-5) is:

y-(-5)=16(x-3).

I'm going to solve this for y.

y-(-5)=16(x-3)

y+5=16(x-3)

Subtract 5 on both sides:

y=16(x-3)-5

What this means is for values x near x=3 is that:

g(x) \approx 16(x-3)-5.

Let's evaluate this approximation function for g(2.9).

g(2.9) \approx 16(2.9-3)-5

g(2.9) \approx 16(-.1)-5

g(2.9) \approx -1.6-5

g(2.9) \approx -6.6

Let's evaluate this approximation function for g(3.1).

g(3.1) \approx 16(3.1-3)-5

g(3.1) \approx 16(.1)-5

g(3.1) \approx 1.6-5

g(3.1) \approx -3.4

b) To determine if these are over approximations or under approximations I will require the second derivative.

If g'' is positive, then it leads to underestimation (since the curve is concave up at that number).

If g'' is negative, then it leads to overestimation (since the curve is concave down at that number).

g'(x)=x^2+7

g''(x)=2x+0

g''(x)=2x

2x is positive for x>0.

2x is negative for x.

That is, g''(2.9)>0 \text{ and } g''(3.1)>0.

So 2x is positive for both values of x which means that the values we found in part (a) are underestimations.

6 0
4 years ago
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