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3 years ago

What is the measure of an arc intercepted by an inscribed angle whose measure is 88 degrees

Mathematics

1 answer:

I am Lyosha [343]3 years ago

7 0

Answer:

$176\°$

Step-by-step explanation:

we know that

The inscribed angle is half that of the arc it comprises.

In this problem

Let

x-----> the measure of an arc intercepted by an inscribed angle

$88\°=\frac{1}{2}(x)$

$x=2(88\°)=176\°$

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Julien's parents want $64,000 at the end of 3 years. They put their money in an account that yields 10% per year compounded mont

Leto [7]

Answer: $\$47,471$

Step-by-step explanation:

Given

Julien's parents want $64,000 at the end of 3 years

The rate of interest is 10% annually

Suppose they invested P amount initially

So, compound interest, compounded monthly is

$\Rightarrow A=P\left( 1+\frac{r}{n}\right)^{nt}\quad [\text{n=Numner of compounding periods}]\\\\\text{Insert the values}\\\Rightarrow 64,000=P\left( 1+\frac{0.10}{12}\right)^{12\times 3}\\\\\Rightarrow 64,000=P\left( 1+\frac{10}{12}\right)^{36}\\\\\Rightarrow P=\dfrac{64000}{1.3481}\\\\\Rightarrow P=\$\ 47,471.37\approx \$ 47,471$

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3 years ago

Mr. See weights his grades. Tests are worth 60%, quizzes are worth 25% and homework is 15%. Joey's test average is an 82, his qu

goldenfox [79]

Answer:

83.15

Step-by-step explanation:

Joey's current grade can be determined by weighting his scores with the weights given and adding the results

Tests = 0.6 x 82 = 49.2

Quizzes = 0.25 x 77 = 19.25

homework = 0.15 x 98 = 14.7

49.2 + 19.25 + 14.7 = 83.15

4 0

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

Please help and show work

Sunny_sXe [5.5K]

Answer:

Just do scan and solve by brainly really helpful ngl

Step-by-step explanation:

5 0

2 years ago

What is the area of a trapezoid? 47,45,65

Masteriza [31]

Area = 2,520m^2
(47+65)/2 = 56
56x45 = 2,520

8 0

2 years ago