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2 years ago

Use the propotion to convert millimeters to centimeters.

Mathematics

2 answers:

aksik [14]2 years ago

8 0

Number 1 is 4.7cm
number 2 is 720 cm

PilotLPTM [1.2K]2 years ago

6 0

Answer:

Step-by-step explanation:

Since 10 mm = 1 cm

we can convert 47 mm as 47x1/10 = 4.7 cm.

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SInce conversion factor for milligrams to grams is 1000 mg = 1 gram

we have

2900 mg = 2900/1000 = 2.9 gms.

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For conversion of m to cm we have conversion factors as

1 m = 100 cm.

Hence 7.2 m = 7.2x100 cm = 720 cm.

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Solve the simultaneous equation 2x-5y=9 3x+4y=2

Dmitriy789 [7]

Answer/Step-by-step explanation:

2x-5y=9

Add 5y to both sides of the equation.

2x=9+5y

Divide each term by 2 and simplify.

Divide each term in 2x=9+5y by 2

$\frac{2x}{2} =\frac{9}{2} +\frac{5y}{2}$

Cancel the common factor of 2.

Divide x by 1.

$x=\frac{9}{2} +\frac{5y}{2}$

$y=\frac{-9}{5} +\frac{2x}{5}$

----------------------------------------------------------------------------------------------------------------

3x+4y=2

Add 4y to both sides of the equation

3x=2+4y

Divide each term by 3 and simplify.

Divide each term in 3x=2+4y by 3.

$\frac{3x}{3} =\frac{2}{3} +\frac{4y}{3}$

Cancel the common factor of 3.

Divide x by 1.

$x=\frac{2}{3} +\frac{4y}{3}$

$y=\frac{-1}{2} +\frac{3x}{4}$

5 0

2 years ago

-7πh = 98π solve for h

ICE Princess25 [194]

Answer:

h = -14

Step-by-step explanation:

-7πh = 98π

⇔ -7h = 98 (Just Divide both sides by π)

⇔ h = 98 ÷ (-7) = -14

5 0

7 months ago

Read 2 more answers

QUESTION 1 The driving distance for the top 100 golfers on the PGA tour is between 284.7 and 310.6 yards.Assume that the driving

GarryVolchara [31]

Answer:

$P(X>300)$

And we can find this probability with the complement rule:

$P(X>300)= 1-P(X$

Step-by-step explanation:

For this case we define the random variable X ="driving distance for the top 100 golfers on the PGA tour" and we know that:

$X \sim Unif (a=284.7, b=310.6)$

And for this case the probability density function is given by:

$f(x) =\frac{1}{310.6 -284.7} =0.0386 , 284.7 \leq X \leq 310.6$

And the cumulative distribution function is given by:

$F(x) =\frac{x-284.7}{310.6-284.7} , 284.7 \leq X \leq 310.6$

And we want to find this probability:

$P(X>300)$

And we can find this probability with the complement rule:

$P(X>300)= 1-P(X$

6 0

3 years ago

The centers for disease control and prevention reported that 25% of baby boys 6-8 months old in the united states weigh more tha

melisa1 [442]

Answer:

0.1971 ( approx )

Step-by-step explanation:

Let X represents the event of weighing more than 20 pounds,

Since, the binomial distribution formula is,

$P(x)=^nC_r p^r q^{n-r}$

Where, $^nC_r=\frac{n!}{r!(n-r)!}$

Given,

The probability of weighing more than 20 pounds, p = 25% = 0.25,

⇒ The probability of not weighing more than 20 pounds, q = 1-p = 0.75

Total number of samples, n = 16,

Hence, the probability that fewer than 3 weigh more than 20 pounds,

$P(X$

$=^{16}C_0 (0.25)^0 (0.75)^{16-0}+^{16}C_1 (0.25)^1 (0.75)^{16-1}+^{16}C_2 (0.25)^2 (0.75)^{16-2}$

$=(0.75)^{16}+16(0.25)(0.75)^{15}+120(0.25)^2(0.75)^{14}$

$=0.1971110499$

$\approx 0.1971$

3 0

2 years ago

If perpendiculars from any point within an angle on its arms are equal, prove that it lies on the bisector of that angle

Oxana [17]

Your question can be quite confusing, but I think the gist of the question when paraphrased is: Prove that the perpendiculars drawn from any point within the angle are equal if it lies on the angle bisector?

Please refer to the picture attached as a guide you through the steps of the proofs. First. construct any angle like ∠ABC. Next, construct an angle bisector. This is the line segment that starts from the vertex of an angle, and extends outwards such that it divides the angle into two equal parts. That would be line segment AD. Now, construct perpendicular line from the end of the angle bisector to the two other arms of the angle. This lines should form a right angle as denoted by the squares which means 90° angles. As you can see, you formed two triangles: ΔABD and ΔADC. They have congruent angles α and β as formed by the angle bisector. Then, the two right angles are also congruent. The common side AD is also congruent with respect to each of the triangles. Therefore, by Angle-Angle-Side or AAS postulate, the two triangles are congruent. That means that perpendiculars drawn from any point within the angle are equal when it lies on the angle bisector

4 0

3 years ago