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monitta
2 years ago
11

Steven recorded the growth of a plant over 10 weeks for his science project. He made a graph that shows how much the plant grew

each week. The plant was 2 in. tall when Steven started his project. At week 5, the plant was 8 in. tall.
Which equation represents Steven’s situation, where x is the number of weeks and y is the height of the plant?

y=85x+2
y=65x+2
y=56x+2
y=58x+2
Mathematics
2 answers:
Margarita [4]2 years ago
4 0

Answer:

It's y=6/5x+2


Olenka [21]2 years ago
3 0
The answer is y=56x+1.
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David made a class banner out of a large rectangular piece of paper. He cut a
Irina18 [472]

Answer:

100 in²

Step-by-step explanation:

The area of the banner is equal to the area of the initial rectangle minus the area of the cutout triangle.

The rectangle has a height of 8 inches and width of 14 inches, so its area is:

A = (8 in) (14 in) = 112 in²

The triangle has a base of 8 inches and a height of 3 inches, so its area is:

A = ½ (8 in) (3 in) = 12 in²

So the area of the banner is 112 in² − 12 in² = 100 in².

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2 years ago
Given ΔMNO, find the measure of ∠LMN.
Novay_Z [31]

Answer:

  104°

Step-by-step explanation:

If segments NO and NM are congruent, then angles NMO and NOM are congruent. So, their supplements, angles NML and NOP are congruent. That is ...

  ∠NML ≅ ∠NOP = 104°

  ∠NML = 104°

3 0
1 year ago
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Sunny earns $12 per hour delivering cakes. She worked for x hours this week. Unfortunately, she was charged $15 for a late deliv
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$12 times X - 15

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8 0
2 years ago
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In △ABC, point P∈ AB is so that AP:BP=1:3 and point M is the midpoint of segment CP . Find the area of △ABC if the area of △BMP
ddd [48]

M is mid point of CP. M will divide the \Delta BPC in two equal parts \Delta BMC and    \Delta BMP.

Area of \Delta BMP is equal to 21m^2

Since, \Delta BMC = \Delta BMP

Area of  \Delta BPC = Area of  \Delta BMC +Area of \Delta BMP =  21 + 21 = 42m^2

and since ratio of AP:BP =1:3  so the area of \Delta BMP will be 1/3 of Area of \Delta ABC

hence, Area of \Delta ABC = 63m^2

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1 year ago
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1 point) As reported in "Runner's World" magazine, the times of the finishers in the New York City 10 km run are normally distri
BigorU [14]

Answer:

(a) E (X) = 61 and SD (X) = 9

(b) E (Z) = 0 and SD (Z) = 1

Step-by-step explanation:

The time of the finishers in the New York City 10 km run are normally distributed with a mean,<em>μ</em> = 61 minutes and a standard deviation, <em>σ</em> = 9 minutes.

(a)

The random variable <em>X</em> is defined as the finishing time for the finishers.

Then the expected value of <em>X</em> is:

<em>E </em>(<em>X</em>) = 61 minutes

The variance of the random variable <em>X</em> is:

<em>V</em> (<em>X</em>) = (9 minutes)²

Then the standard deviation of the random variable <em>X</em> is:

<em>SD</em> (<em>X</em>) = 9 minutes

(b)

The random variable <em>Z</em> is the standardized form of the random variable <em>X</em>.

It is defined as:Z=\frac{X-\mu}{\sigma}

Compute the expected value of <em>Z</em> as follows:

E(Z)=E[\frac{X-\mu}{\sigma}]\\=\frac{E(X)-\mu}{\sigma}\\=\frac{61-61}{9}\\=0

The mean of <em>Z</em> is 0.

Compute the variance of <em>Z</em> as follows:

V(Z)=V[\frac{X-\mu}{\sigma}]\\=\frac{V(X)+V(\mu)}{\sigma^{2}}\\=\frac{V(X)}{\sigma^{2}}\\=\frac{9^{2}}{9^{2}}\\=1

The variance of <em>Z</em> is 1.

So the standard deviation is 1.

8 0
1 year ago
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