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3 years ago

Find the value of x: x/6 = 16/48

Mathematics

1 answer:

Paul [167]3 years ago

6 0

Answer:

x = 2

Step-by-step explanation:

x/6 = 16/48

48x = 6 . 16

48x = 96

x = 96/48

x = 2

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Simplify (x^6)^2 • x^3

givi [52]

Answer:

x^15

Step-by-step explanation:

Recall these rules of exponents:

(a^m)^n = a^mn

a^m • a^n = a^(m + n)

(x^6)² • x³ = x^(2 • 6) • x³ = x^12 • x³ = x^(12 + 3) = x^15

8 0

3 years ago

Can't somebody help me math work please i need help

valentina_108 [34]

Answer:

TUP and PUQ

Step-by-step explanation:

Adjacent angles share a common vertex and common side, but don't overlap.

7 0

3 years ago

What are the odds against choosing a red marble from a bag that contains two blue marbles, one green marble, seven white marbles

Alina [70]

Answer:

4/14

Step-by-step explanation:

2+1+7+4=14

red marbles are 4/14

7 0

3 years ago

Read 2 more answers

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

3 years ago

3.<br> * + 4y = 6<br> = -x +3<br> What is the answer

miv72 [106K]

Answer:

the answers is -4

Step-by-step explanation:

-1+3 I think this the answer

6 0

4 years ago