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Andreas93 [3]
3 years ago
13

What are the first 3 common multiples of 6 and 9

Mathematics
1 answer:
ehidna [41]3 years ago
3 0
18,36, and 54. These are the common multiple of 6 and 9
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Which equation represents the vertical line passing through (7,-3)
defon

Answer:

B- y=-3

Step-by-step explanation:

4 0
3 years ago
The following are the temperatures of 10 days in winter in °C. − 3 , 0 , 10 , 2 , 8 , − 5 , − 1 , 9 , 2 , − 3 Work out the mean
marissa [1.9K]

The mean temperature of the given data − 3 , 0 , 10 , 2 , 8 , − 5 , − 1 , 9 , 2 , − 3 is 1.9

mean= 1.9

<h3>What is mean of the data?</h3>

The mean, or average, is calculated by adding up the data and dividing by total number of data.

Given data: − 3 , 0 , 10 , 2 , 8 , − 5 , − 1 , 9 , 2 , − 3

Now, mean =   \frac{sum \;of\; observation}{Total \;number \;of \;observation}

mean = \frac{(-3)+ 0+10+2+8+(-5)+ (-1)+ 9+2+(-3)}{10}

           = \frac{19}{10}

mean= 1.9

Hence, the mean of − 3 , 0 , 10 , 2 , 8 , − 5 , − 1 , 9 , 2 , − 3  is 1.9

Learn more about mean here:

brainly.com/question/22871228

#SPJ1

3 0
2 years ago
If 10% of men are bald, what is the probability that fewer than 100 in a random sample of 818 men are bald? (Answers must be in
Greeley [361]

Answer: the probability that fewer than 100 in a random sample of 818 men are bald is 0.9830

Step-by-step explanation:

Given that;

p = 10% = 0.1

so let q = 1 - p = 1 - 0.1 = 0.9

n = 818

μ = np = 818 × 0.1 = 81.8

α = √(npq) = √( 818 × 0.1 × 0.9 ) = √73.62 = 8.58

Now to find P( x < 100)

we say;

Z = (X-μ / α) = ((100-81.8) / 8.58) = 18.2 / 8.58 = 2.12

P(x<100) = P(z < 2.12)

from z-score table

P(z < 2.12) = 0.9830

Therefore the probability that fewer than 100 in a random sample of 818 men are bald is 0.9830

4 0
3 years ago
It is generally claimed that the average body temperature for healthy human adults is 98.6°F. To test this claim a simple random
Rzqust [24]

Answer:

Explained below.

Step-by-step explanation:

The information provided is as follows:

\mu=98.6^{o}F\\\bar x=98.1^{o}F\\s=0.9^{o}F\\n=9

(1)

A single mean test is to be performed in this case.

As the population standard deviation is not provided, a one-sample <em>t</em>-test will be used.

The correct option is b.

(2)

The null hypothesis is:

<em>H</em>₀: The average temperature in the population is 98.6°F, i.e. <em>μ </em>= 98.6°F.

The correct option is b.

(3)

The alternative hypothesis is:

<em>Hₐ</em>: The average temperature in the population is less than 98.6°F, i.e. <em>μ </em>< 98.6°F.

The correct option is c.

(4)

The standard deviation of the sample mean is as follows:

SD_{\bar x}=\frac{s}{\sqrt{n}}=\frac{0.9}{\sqrt{9}}=0.3^{o}F

Thus, the value of SD is 0.3°F.

(5)

Compute the value of test statistic as follows:

t=\frac{\bar x-\mu}{SD_{\bar x}}

  =\frac{98.1-98.6}{0.3}\\\\=-1.666666667\\\\\approx -1.67

Thus, the value of test statistic is -1.67.

(6)

The degrees of freedom of the test are:

df = n - 1

   = 9 - 1

   = 8

Thus, the degrees of freedom of the test is 8.

8 0
3 years ago
84 ÷ (17 × 3 - 23) × 4
Zielflug [23.3K]

Answer: 12

Step-by-step explanation:

4 0
3 years ago
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