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skad [1K]
3 years ago
13

What is a triangle inscribed in a semicircle called

Mathematics
1 answer:
Tanya [424]3 years ago
5 0
Right traingle

I hope it helped you!
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Is percent of change less than 100
Ivanshal [37]
You can't change more than 100 percent
6 0
3 years ago
20 - 12/x = 16 ; x = 3
uranmaximum [27]

Answer: 20−12/3=?16

16=16

True

Step-by-step explanation: hope this help:)

3 0
3 years ago
Read 2 more answers
In 7th grade there are 58 students with red-hair and 609 students with another color hair. The numbers of red-haired students in
mr_godi [17]

In 7th grade:

red-hair students =58

another color hair students =609

In 8th grade:

red-hair students =60

Let's assume

another color hair students =x

now, we are given

The numbers of red-haired students in 8th grade are proportional to the numbers in 7th

so,

\frac{60}{58} =\frac{x}{609}

now, we can solve for x

we get

x=609*\frac{60}{58}

we get

x=630

so, students have another color hair in 8th grade is 630.........Answer

5 0
3 years ago
A manufacturing process produces semiconductor chips with a known failure rate of . If a random sample of chips is selected, app
AleksandrR [38]

Answer:

The probability that at least 14 of the chips will be defective is 0.6664.

Step-by-step explanation:

The complete question is:

A manufacturing process produces semiconductor chips with a known failure rate of 5.4%. If a random sample of 300 chips is selected, approximate the probability that at least 14 will be defective. Use the normal approximation to the binomial with a correction for continuity .

Solution:

Let <em>X</em> = number of defective chips.

The probability that a chip is defective is, <em>p</em> = 0.054.

A random sample of <em>n</em> = 300 chips is selected.

A chip is defective or not is independent of the other chips.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 300 and <em>p</em> = 0.054.

But the sample selected is too large.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

 np=300\times 0.054=16.2>10\\n(1-p)=300\times (1-0.054)=283.8>10

Thus, a Normal approximation to binomial can be applied.

So,  X\sim N(\mu =16.2,\ \sigma^{2}=15.3252).

Compute the probability that at least 14 of the 300 chips will be defective as follows:

Use continuity correction:

P (X ≥ 14) = P (X > 14 + 0.50)

               = P (X > 14.50)

               =P(\frac{X-\mu}{\sigma}>\frac{14.50-16.20}{\sqrt{15.3252}})

                =P(Z>-0.43)\\=P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that at least 14 of the chips will be defective is 0.6664.

5 0
3 years ago
I’ll mark you brainliest
levacccp [35]

Answer:

42

Step-by-step explanation:

In a rectangle, the diagonals have the same length. Therefore:

3x+6=5x-18 \\\\3x+24=5x \\\\ 24=2x \\\\x=12 \\\\BD=5(12)-18=60-18=42

Hope this helps!

7 0
3 years ago
Read 2 more answers
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