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2 years ago

Translate into an algebraic expression, find the number if 10% of it is m

Mathematics

1 answer:

I am Lyosha [343]2 years ago

5 0

Answer:

m/0.1

Step-by-step explanation:

we want to find it

-----

10%*x=m

OF means multiply

IT is any variable

IS is equal

10%*x=m

0.1*x=m

x=m/0.1 IN THE ANSWER DO NOT INCLUDE x

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The sample consisted of 50 night students, with a sample mean GPA of 3.02 and a standard deviation of 0.08, and 25 day students,

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Answer: The test statistic is t= -0.90.

Step-by-step explanation:

Since we have given that

n₁ = 50

n₂ = 25

$\bar{x_1}=3.02\\\\\bar{x_2}=3.04\\\\\sigma_1=0.08\\\\\sigma_2=0.06$

So, s would be

$s=\sqrt{\dfrac{n_1\sigma_1^2+n_2\sigma_2^2}{n_1+n_2-2}}\\\\s=\sqrt{\dfrac{50\times 0.08^2+25\times 0.06^2}{50+25-2}}\\\\s=0.075$

So, the value of test statistic would be

$t=\dfrac{\bar{x_1}-\bar{x_2}}{s(\dfrac{1}{n_1}+\dfrac{1}{n_2})}\\t=\dfrac{3.02-3.04}{0.074(\dfrac{1}{50}+\dfrac{1}{25})}\\\\t=\dfrac{-0.04}{0.074(0.02+0.04)}\\\\t=\dfrac{-0.04}{0.044}\\\\t=-0.90$

Hence, the test statistic is t= -0.90.

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2 years ago

Sherman has 4 cats and 1 dog. He wants to buy a toy for each of his pets. Sherman has $46 to spend on

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Step-by-step explanation:

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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)

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