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lawyer [7]
3 years ago
14

Write each percent as a decimal a 80% B 3% C 12.5% D 125%​

Mathematics
2 answers:
notsponge [240]3 years ago
5 0

Answer:

a. 0.8

b 0.03

c. 0.125

d. 1.25

Tema [17]3 years ago
4 0

Answer:

a .8

b .03

c .125

d 125.0

Step-by-step explanation:

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3a² + 11a – 42<br><br>this is factoring in algebra 2 (high school) <br>​
N76 [4]

To factor,

<h2>[[[</h2>

1) First multiply coefficient of a² and constant no,

That is,

3×(-42)=-126

Since the<u> resultant no is negative</u>, you should find two such factors of 126 <u>which</u> <u>will give us the coefficient of a (=11)</u> on subracting those factors.

2) Find the factor

126=2×3×3×7

=18×7

18 and 17 are factors of 126

Also,18-7 =11.

So they are required factors for factoring,

<h2>]]]</h2>

Once you have understood above steps you can solve on your own. All you need to do is split 11 into factors ,take common terms and you will get answer.

<u>Answer:</u>

3a²+11a-42

=3a²+(18-7)a -42

=3a²+18a-7a-42

=3a(a+6) -7(a+6)

=(a+6)(3a-7)

3 0
3 years ago
Mrs. Baker conducted a survey in her classroom to determine what the students preferred to do during the summer break. Her surve
navik [9.2K]

Answer:

B. 840

Step-by-step explanation:

21 is 70% of 30 and 70% of 1,200 is 840.

4 0
3 years ago
Read 2 more answers
Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample  means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean  is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})

                           =P(Z>2.63)\\=1-P(Z

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

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

8 0
3 years ago
Elaina wrote two correct variable expressions to represent the phrase “one-fourth of a number.” Which variable expressions could
ivolga24 [154]
The answer is n over 4 
6 0
3 years ago
What is the most appropriate SI unit for measuring the following items?
pochemuha

Answer: (d) meter

have a good day :)

5 0
3 years ago
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