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3 years ago

Find the missing terms. 3, _, _, 21 (arithmetic)

Mathematics

1 answer:

natka813 [3]3 years ago

6 0

Answer:

3, 9, 15, 21

Step-by-step explanation:

3x1=3 v

3x2=6

3x3=9 v

3x4=12

3x5=15 v

3x6=18

3x7=21 v

Skip one each time, 3, 9, 15, 21

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Step-by-step explanation:

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3 years ago

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Please answer this question

scZoUnD [109]

Domain in numerical order: {-7, -1, 0, 1}

Range in numerical order: {-1, 0, 1, 2, 7}

Which number is not part of the range of this relation? = -7

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3 years ago

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SIZIF [17.4K]

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3 years ago

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Solve the following system of equations. What is the y value of the solution?

Svet_ta [14]

Answer:

y = 2

Step-by-step explanation:

x + 5y - 10 = 0

X = 4y - 8

since x is given as = 4y - 8 we can use this in first equation

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4 years ago

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We wish to give a 90% confidence interval for the mean value of a normally distributed random variable. We obtain a simple rando

coldgirl [10]

Answer: $(9.27025,\ 11.12975)$

Step-by-step explanation:

Given : Sample size : n= 9

Degree of freedom = df =n-1 =8

Sample mean : $\overline{x}=10.2$

sample standard deviation : $s= 1.5$

Significance level ; $\alpha= 1-0.90=0.10$

Since population standard deviation is not given , so we use t- test.

Using t-distribution table , we have

Critical value = $t_{\alpha/2, df}=t_{0.05 , 8}=1.8595$

Confidence interval for the population mean :

$\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}$

90% confidence interval for the mean value will be :

$10.2\pm (1.8595)\dfrac{1.5}{\sqrt{9}}$

$10.2\pm (1.8595)\dfrac{1.5}{3}$

$10.2\pm (1.8595)(0.5)$

$10.2\pm (0.92975)$

$(10.2-0.92975,\ 10.2+0.92975)$

$(9.27025,\ 11.12975)$

Hence, the 90% confidence interval for the mean value= $(9.27025,\ 11.12975)$

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3 years ago

Find the missing terms. 3, _____, _____, 21 (arithmetic)

Find the missing terms. 3, _, _, 21 (arithmetic)