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3 years ago

What ratio is equivalent to 2/8

Mathematics

1 answer:

stiv31 [10]3 years ago

5 0

$\huge\text{Hey there!}$

$\mathsf{\dfrac{2}{8}=\dfrac{2\div2}{8\div2}=\boxed{\bf{\dfrac{1}{4}}}}}$

$\mathsf{Rewrite\ \dfrac{1}{4}\ as\ ratio\ and\ that\ will\ be\ your\ answer!}$

$\mathsf{\dfrac{1}{4}=\bf{1:4}}$

$\boxed{\boxed{\bf{Your\ answer: \boxed{\huge\text{\bf{1:4}}}}}}}}\huge\checkmark$

$\large\text{Good luck on your assignment and enjoy your day!}\\\\\dfrac{\frak{LoveYourselfFirst}}{:)}$

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Please I need some answers

Ilia_Sergeevich [38]

-6 x -4 = 24
24/-6 = -4
24/-4 = -6

Hope this helps :)

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3 years ago

Can someone pls help solve this

Len [333]

Answer:

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Step-by-step explanation:

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3 years ago

Carey earns $9.75 working part time on weekends. The table below shows the amount, a, Carey earns for working h hours.

shutvik [7]

$29.25 in 3 hours. explanation: if he makes $9.75 in 1 hour then 9.75 x 3 (hours) would equal $29.25

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3 years ago

A survey of 700 adults from a certain region asked, "What do you buy from your mobile device?" The results indicated that 59%

Kryger [21]

Answer:

a) the test statistic z = 1.891

the null hypothesis accepted at 95% level of significance

b) the critical values of 95% level of significance is zα =1.96

c) 95% of confidence intervals are (0.523 ,0.596)

Step-by-step explanation:

A survey of 700 adults from a certain region

Given sample sizes $n_{1} = 400 and n_{2} = 300$

Proportion of mean $p_{1} = \frac{236}{400} = 0.59 and p_{2} = \frac{156}{300} = 0.52$

Null hypothesis H0 : assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

Alternative hypothesis H1:- p1 ≠ p2

a) The test statistic is

$Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )} } }$

where $p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}$

on calculation we get p = 0.56

now q =1-p = 1-0.56=0.44

$Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )} } }\\ =\frac{0.56-0.52}{\sqrt{0.56X0.44}(\frac{1}{400}+\frac{1}{300} }$

after calculation we get z = 1.891

b) The critical value at 95% confidence interval zα = 1.96 (from z-table)

The calculated z- value < the tabulated value

therefore the null hypothesis accepted

conclusion:-

assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

c) 95% confidence intervals

The confidence intervals are P± 1.96(√PQ/n)

we know that = $p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}$

after calculation we get P = 0.56 and Q =1-P =0.44

Confidence intervals are ( P- 1.96(√PQ/n), P+ 1.96(√PQ/n))

now substitute values , we get

( 0.56- 1.96(√0.56X0.44/700), 0.56+ 1.96(0.56X0.44/700))

on simplification we get (0.523 ,0.596)

Therefore the population proportion (0.56) lies in between the 95% of confidence intervals (0.523 ,0.596)

3 0

3 years ago

11/7 in a mixed fraction

Mama L [17]

7 goes into 11 one time, which leaves you with 1 4/7.

6 0

3 years ago

Read 2 more answers