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2 years ago

On April 1, Preston McCord deposited $1,400 in a savings account that pays annual interest of 3.2%

Mathematics

1 answer:

lesya692 [45]2 years ago

4 0

Answer:

After 6 months, the interest generated by the investment would be $21.10.

Step-by-step explanation:

To determine the interest that Preston McCord could earn by investing $ 1,400 in an account that pays 3.2% annual interest compounded monthly, leaving said money invested for a period of 6 months, it is necessary to perform the following calculation:

X = 1,400 x (1 + 0.032 / 6) ^ 0.5x6

X = 1,421.10

1421.10 - 1400 = 21.10

Thus, after 6 months, the interest generated by the account will be $ 21.10.

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Step-by-step explanation:

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If L, M and N are midpoints of the sides of PQR, PR = 46, PQ = 40, and LN=17, find each measure

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Check the picture below.

let's recall that the midsegment of a triangle is the segment that's half-way to both endsides and is at the same time parallel to the 3rd side, is also half the length of the parallel side, so-called the base.

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If sin theta = 4/5 and theta is in quadrant 2, the value of cot theta is

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Answer:

Cot(theta) = - 0.75 or -3/4

Step-by-step explanation:

The hypotenuse is 5

The y value is 4

We need to find the corresponding x value.

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X = ?

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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%

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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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