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3 years ago

Jenna knits scarves and then sells them on Etsy, an online marketplace. Let C(x) = 4x + 20 represent the cost C in

Mathematics

1 answer:

Alex Ar [27]3 years ago

3 0

Answer:

Step-by-step explanation:

We are given that Jenna knits scarves and sells them on Etsy an online market place.

$C(x)=4x+20$

Where C=Represents the cost in dollars

x=Number of scarves

We have to create a table to show the relationship between the number of scarves x and the cost C

Substitute x=1

$C(1)=4(1)+20=$ $24

Substitute x=2

$C(2)=4(2)+20=$ $28

$C(3)=4(3)+20=$ $32

$C(4)=4(4)+20=$ $36

$C(5)=4(5)+20=$ $40

$C(6)=4(6)+20=$ $44

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GalinKa [24]

Answer:

true

Step-by-step explanation:

the lines of the new shape are parallel to the lines of the original shape, there seems to be the same scaling factor for all sides and all "projection" lines BB' and CC' cross at A, which is also A' Ave the center of the dilation.

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2 years ago

How do I solve 1/10x+5=-3x-13+4x ?

Temka [501]

Answer:

x = 20

I am not sure if you wanted the answer. Sorry!

Step-by-step explanation:

Let's solve your equation step-by-step.

1/10 x + 5 = - 3x - 13 + 4x

Step 1: Simplify both sides of the equation.

1/10 x + 5 = - 3x - 13 + 4x

1/10 x + 5 = - 3x + - 13 + 4x

1/10 x + 5 = (- 3x + 4x) + (- 13)

1/10 x + 5 = x + - 13

1/10 x + 5 = x - 13

Step 2: Subtract x from both sides.

1/10 x + 5 - x = x - 13 - x

- 9/10 x + 5 = - 13

Step 3: Subtract 5 from both sides

- 9/10 x + 5 - 5 = - 13 - 5

- 9/10 x = - 18

Step 4: Multiply both sides by 10/(- 9).

(10/ - 9) * (- 9/10 x) = (01/ - 9) * (- 18)

x = 20

Hope this helps!

4 0

3 years ago

Read 2 more answers

Step 3a: Calculating the average rate

Alex73 [517]

9514 1404 393

Answer:

a) average rate = (total distance)/(total time)

b) Rave = 2·R1·R2/(R1 +R2)

c) cheetah's average rate ≈ 50.91 mph

Step-by-step explanation:

a) Let AB represent the distance from A to B. Let t1 and t2 represent the travel times (in hours) on leg1 and leg2 of the trip, respectively. Then the distances traveled are...

First leg distance: AB = 70·t1 ⇒ t1 = AB/70

Second leg distance: AB = 40·t2 ⇒ t2 = AB/40

The average rate is the ratio of total distance to total time:

average rate = (AB +AB)/(t1 +t2)

average rate = 2AB/(AB/70 +AB/40) = 2/(1/70 +1/40) = 2(40)(70)/(70+40)

average rate = 560/11 = 50 10/11 . . . mph

No equations are given, so we cannot compare what we wrote with the given equations. In each step of the solution, we have used the rules of algebra and equality.

b) For two rates over the same distance (as above), the average is their harmonic mean:

average rate = 2r1·r2/(r1+r2)

c) The cheetah's average rate was 50 10/11 mph ≈ 50.91 mph.

4 0

3 years ago

What is the solution?

monitta

Answer:

(4, 4)

Step-by-step explanation:

y = 7/4 x - 3 graph (0, -3) (2, .5), (4, 4) (8, 11). Draw the line

y = x graph (0,0) (2, 2), (4, 4) , ( -2, -2). Draw the line

The two lines intersect at point (4, 4)

8 0

3 years ago

The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

yulyashka [42]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.5 \ in$

Generally the Null hypothesis is $H_o : \mu = 0. 5 \ in$

The Alternative hypothesis is $H_a : \mu \ne 0.5 \ in$

Considering the parameter given for part a

The sample size is n = 15

The test statistics is t = 1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = 2.145$

We are making use of this $t_{\frac{\alpha }{2} }$ because it is a one-tail test

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that $t < t_{\frac{\alpha }{2} }$ so the null hypothesis would not be rejected

Considering the parameter given for part b

The sample size is n = 15

The test statistics is t = -1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = -2.145$

Looking at the value of t and $t_{\frac{\alpha}{2} ,df }$ the we see that t does not lie in the area covered by $t_{\frac{\alpha}{2} , df }$ (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part c

The sample size is n = 26

The test statistics is t = -2.55

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = 2.787$

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that t does not lie in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part d

The sample size is n = 26

The test statistics is t = -3.95

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = -2.787$

Looking at the value of t and $t_{\frac{\alpha}{2} }$ the we see that t lies in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we reject the null hypothesis

6 0

3 years ago