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2 years ago

The graph shows the cost (C) of shipping a package based on its mass in pounds (p)

Mathematics

1 answer:

EleoNora [17]2 years ago

5 0

Answer:

C = 4p

Step-by-step explanation:

Because it starts at 0 pounds costs 0 dollars, we know it starts at the origin and has no y-intercept. Because it rises up 20 dollars for every 5 pounds, we can use rise/run to do 20/5 to get an increase of 4 dollars per pound.

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I need help on number 4

Svetach [21]

Ok
Part B: Both of the children are incorrect. The children have the same amount of money
Explanation: they have the same amount because 6 dimes are 60 cents... Hannah says she has 60 cents while Mason says he has 6 dimes. Got it?
Part A: 6 dimes = 60 cents
Explanation: None

6 0

4 years ago

The ratio of dogs to cats in a pet store is 4 : 34:3 . The ratio of cats to fish in the same pet store is 2 : 92:9. If there ar

dybincka [34]

Answer: In the pet store, there are 81 fish.

Step-by-step explanation:

We know that:

The ratio of dogs to cats is 4:3

(this means that for every 4 dogs, there are 3 cats)

The ratio of cats to fish is 2:9

(this means that for every 2 cats, there are 9 fish).

Now, we know that there are a total of 24 dogs in the pet store.

Remembering that for each 4 dogs we have 3 cats, we first need to see how many times we have 4 dogs in a total of 24 dogs, this is:

24/4 = 6

So we have 6 times 4 dogs, then we will have 6 times 3 cats, this is a total of:

6*3 cats = 18 cats.

And now we know that for every 2 cats we have 9 fish.

Then we need to see how many times we have 2 cats in a total of 18 cats, this is:

18/2 = 9.

And for each one of these 9 groups, we will have 9 fish, then the total number of fish is:

9*9 = 81 fish

In the pet store, there are 81 fish.

6 0

3 years ago

Find the equation of the line passing through the following point.

Valentin [98]

$y+0=1*(x+0)$ or $y=x$

7 0

3 years ago

When it is operating properly, a chemical plant has a daily production rate that is normally distributed with a mean of 885 tons

-BARSIC- [3]

Answer:

The test statistic Z = 1.844 < 1.96 at 0.05 level of significance

Null hypothesis is accepted

Yes he is right

The manager claims that at least 95 % probability that the plant is operating properly

Step-by-step explanation:

Explanation:-

Given data Population mean

μ = 885 tons /day

Given random sample size

n = 60

mean of the sample

x⁻ = 875 tons/day

The standard deviation of the Population

σ = 42 tons/day

Null hypothesis:- H₀: The manager claims that at least 95 % probability that the plant is operating properly

Alternative Hypothesis :H₁: The manager do not claims that at least 95 % probability that the plant is operating properly

Level of significance = 0.05

The test statistic

$Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{875 -885}{\frac{42}{\sqrt{60} } }$

$Z = \frac{-10}{5.422} = -1.844$

|Z| = |-1.844| = 1.844

The tabulated value

$Z_{\frac{0.05}{2} } = Z_{0.025} = 1.96$

The calculated value Z = 1.844 < 1.96 at 0.05 level of significance

Null hypothesis is accepted

Conclusion:-

The manager claims that at least 95 % probability that the plant is operating properly

8 0

3 years ago

I honestly don’t get how to solve this

ruslelena [56]

Answer:

Two other possible measure ments of AD and AE are

(a) Let AD = 4 units and AE = 6 units.

(b )Let AD = 2 units and AE = 3 units,

Step-by-step explanation:

In ΔABC, given

AB = 8 units, BC = 9 units, AC = 12 units,

D is on AB and E is on AC

Now, when AD = 6 and AE = 9, both ΔABC and ΔADE are similar.

Also, DB = 8 units - 6 units = 2 units,

EC = 12 units - 9 units = 3 units

⇒ $\frac{AD}{DB} = \frac{AE}{EC} = 3$

To adjust the points D and E in such a way, triangles REMAIN SIMILAR.

(a) Let AD = 4 units and AE = 6 units

Then, DB = 8 units - 4 units = 4 units,

EC = 12 units - 6 units = 6 units

⇒ $\frac{AD}{DB} = \frac{4}{4} = 1 , \\ \frac{AE}{EC} = \frac{6}{6} = 1$

⇒ $\frac{AD}{DB} = \frac{AE}{EC} = 1$

Hence, ΔABC and ΔADE are similar.

(b) Let AD = 2 units and AE = 3 units

Then, DB = 8 units - 2 units = 6 units,

EC = 12 units - 3 units = 9 units

⇒ $\frac{AD}{DB} = \frac{2}{6} = \frac{1}{3} , \\ \frac{AE}{EC} = \frac{3}{9} = \frac1}{3}$

⇒ $\frac{AD}{DB} = \frac{AE}{EC} =\frac{1}{3}$

Hence, ΔABC and ΔADE are similar.

4 0

3 years ago