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MA_775_DIABLO [31]
3 years ago
14

What is the scale factor of Figure B to

Mathematics
1 answer:
Kamila [148]3 years ago
5 0
The scale fraction is 2
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0.72 is ten times as much as ________. *
RideAnS [48]

Answer:

0.072 or .072

Step-by-step explanation:

0.72 is ten times as much as 0.072.

0.072 times 10 is .72.

0.72 / 10 = 0.072 .

5 0
3 years ago
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1. Simplify: (-9x3 + 4x2 – 5) - (14x2 - 10x - 3)<br> Show work please
zimovet [89]

Answer:

I got <em>-9x^3-10x^2+10x-2</em>

Step-by-step explanation:

3 0
3 years ago
Brandon is playing hide-and-seek with Katy and Tony. Katy is hiding 63 feet south of Brandon, and Tony is hiding due east of Kat
sergij07 [2.7K]

Answer:

60 feets

Step-by-step explanation:

Given that :

Using Pythagoras rule :

Hypotenus = sqrt(Adjacent^2 + opposite^2)

From the attached picture :

x² = 87^2 - 63^2

x^2 = 7569 - 3969

X^2 = 3600

Take the square root of both sides :

X = sqrt(3600)

X = 60 feets

Hence, Katy and Tony are 60 feets apart

6 0
3 years ago
When solving quadratic equation, what do the X value stand for as it relates to plotting the graph of the equation?
Ber [7]

Step-by-step explanation:

step 1. The x values refer to the domain of the function.

step 2. if you set y = 0 and solve the quadratic equation the x values refer to the point the graph crosses the x axis.

7 0
3 years ago
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
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