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shusha [124]
2 years ago
15

Help plz i dont get it and its very confusing for me and itz due before the 18th and i needa pass or trouble and like i wanna go

od grade so plz help i go to flvs if u also go to flvs come my way plz and help me out
A proportional relationship between the number of pounds of cabbage (x) and the price in dollars (y) is graphed, and the ordered pair (5, 2) is on the graphed line.

Part A: What is the price of 1 pound of cabbage? Show your work. (8 points)
Part B: What does the ordered pair (10, 4) on the graph represent? Explain in words. (2 points)
Mathematics
1 answer:
Oksi-84 [34.3K]2 years ago
8 0

Answer:

A= 0.4 cents per cabbage

B= represents 10 cabbages is $4

Step-by-step explanation:

work for A       2 divided by 5 = 0.4

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mestny [16]

Answer:

swag

Step-by-step explanation:

8 0
3 years ago
A parabola can be drawn given a focus of
Volgvan

Answer:

y=-\frac{1}{4}(x-2)^{2}+9

Step-by-step explanation:

Any point on a given parabola is equidistant from focus and directrix.

Given:

Focus of the parabola is at (2,8).

Directrix of the parabola is y=10.

Let (x,y) be any point on the parabola. Then, from the definition of a parabola,

Distance of (x,y) from focus = Distance of (x,y) from directrix.

Therefore,

\sqrt{(x-2)^{2}+(y-8)^{2}}=|y-10|

Squaring both sides, we get

(x-2)^{2}+(y-8)^{2}=(y-10)^{2}\\(x-2)^{2}=(y-10)^{2}-(y-8)^{2}\\(x-2)^{2}=(y-10+y-8)(y-10-(y-8))...............[\because a^{2}-b^{2}=(a+b)(a-b)]\\(x-2)^{2}=(2y-18)(y-10-y+8)\\(x-2)^{2}=2(y-9)(-2)\\(x-2)^{2}=-4(y-9)\\y-9=-\frac{1}{4}(x-2)^{2}\\y=-\frac{1}{4}(x-2)^{2}+9

Hence, the equation of the parabola is y=-\frac{1}{4}(x-2)^{2}+9.

4 0
3 years ago
Find the point of intersection<br> y=-3x-3<br> y=-3
Alla [95]
The answer would be -3
5 0
2 years ago
Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of
IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

  • np ≥ 10
  • n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.10=1

Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

8 0
3 years ago
The difference between ten times a number and eight times the number is negative ten
IRINA_888 [86]

10n - 8n = -10

Hope this helps!

6 0
3 years ago
Read 2 more answers
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