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4 years ago

Tricia listed her assets and liabilities.

Mathematics

2 answers:

kumpel [21]4 years ago

6 0

Answer:

The correct answer is A. Credit card bill and B. Car loan

Step-by-step explanation:

Let's recall that we consider a personal or familiar liability an obligation to pay money you or your family owe to another person or financial institution like a bank. Liabilities can be classified according to the term, either short such as a credit card or loan from the bank with one-year term, or long such as a car loan or mortgage.

According to that definition, Tricia's liabilities are A. Credit card bill and B. Car loan.

lesantik [10]4 years ago

6 0

Answer:

its a and b

Step-by-step explanation:

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3 years ago

Read 2 more answers

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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3 years ago