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kirill115 [55]
3 years ago
10

The figures in the picture are similar to each other find the value of x

Mathematics
1 answer:
alisha [4.7K]3 years ago
6 0

Answer:

x = 7

Step-by-step explanation:

Similar figures have sides that are proportional.  Setting up a proportion for the sides of the figures will help solve for 'x':

\frac{small}{large}=\frac{3}{6}=\frac{(x-3)}{(x+1)}

Cross-multiply:  3(x + 1) = 6(x - 3)

Distribute:  3x + 3 = 6x - 18

Combine like terms:  21 = 3x

Solve for 'x':  x = 7

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The 2 graphs on the left are exponential
The other 2 are linear
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Plss help me plsss!!​
grin007 [14]

Answer:

  • 20 units

Step-by-step explanation:

The first part is already solved.

<u>We got the scale factor </u>

  • k = 4

<u>Use this to find the missing side:</u>

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3 0
3 years ago
Read 2 more answers
In a health club, research shows that on average, patrons spend an average of 42.5 minutes on the treadmill, with a standard dev
Paha777 [63]

Answer:

0.3114

Option d is right

Step-by-step explanation:

Let X be the time spent on a treadmill in the health club

Given that  research shows that on average, patrons spend an average of 42.5 minutes on the treadmill, with a standard deviation of 5.4 minutes

Also given that X is normal

the probability that randomly selected individual would spent between 30 and 40 minutes on the treadmill.

= P(30

round off to two decimals tog et

the probability that randomly selected individual would spent between 30 and 40 minutes on the treadmill is 0.31

Hence option d is right

3 0
4 years ago
Please Help
Setler79 [48]
The only equation that matches the first point (x, y) = (2, 2) is that of ...

c) y = 3/2x -1
5 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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