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2 years ago

Choose all that apply: Which angle measurements are supplementary to each other?

Mathematics

1 answer:

Lana71 [14]2 years ago

6 0

Answer:

144&36

126&54

21&159

Step-by-step explanation:

Supplementary means angles that add up to 180 degrees

Since these angles add up to 180 degrees, they are therefore supplementary

You might be interested in

In the right ΔABC, AN is the altitude to the hypotenuse. Find BN, AN, and AC, if AB=2√5 in and NC= 1 in.

kipiarov [429]

Answer:

$BN=4\ \text{in}$

$AN=2\ \text{in}$

$AC=\sqrt{5}\ \text{in}$

Step-by-step explanation:

Let $BN=x$

$BC=c$

$NC=y=1\ \text{in}$

$AB=a=2\sqrt{5}\ \text{in}$

$AC=b$

We have the relation

$\dfrac{BC}{AB}=\dfrac{AB}{BN}\\\Rightarrow \dfrac{c}{a}=\dfrac{a}{x}\\\Rightarrow c=\dfrac{a^2}{x}\\\Rightarrow x+1=\dfrac{a^2}{x}\\\Rightarrow x^2+x=20\\\Rightarrow x^2+x-20=0\\\Rightarrow x=\frac{-1\pm \sqrt{1^2-4\times 1\times \left(-20\right)}}{2\times 1}\\\Rightarrow x=4,-5$

$\boldsymbol{BN=4\ \text{in}}$

$h=\sqrt{a^2-x^2}\\\Rightarrow h=\sqrt{(2\sqrt{5})^2-4^2}\\\Rightarrow h=2\ \text{in}$

$\boldsymbol{AN=2\ \text{in}}$

$b=\sqrt{h^2+y^2}\\\Rightarrow b=\sqrt{2^2+1^2}\\\Rightarrow b=\sqrt{5}\ \text{in}$

$\boldsymbol{AC=\sqrt{5}\ \text{in}}$

8 0

2 years ago

A survey was conducted in the United Kingdom, where respondents were asked if they had a university degree. One question asked,

katovenus [111]

Answer:

$z=\frac{0.121-0.0892}{\sqrt{0.101(1-0.101)(\frac{1}{373}+\frac{1}{639})}}=1.620$

$p_v =2*P(Z>1.620)=0.105$

If we compare the p value and using any significance level for example $\alpha=0.05$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the two proportions are not statistically different at 5% of significance

Step-by-step explanation:

Data given and notation

$X_{1}=45$ represent the number of correct answers for university degree holders

$X_{2}=57$ represent the number of correct answers for university non-degree holders

$n_{1}=373$ sample 1 selected

$n_{2}=639$ sample 2 selected

$p_{1}=\frac{45}{373}=0.121$ represent the proportion of correct answers for university degree holders

$p_{2}=\frac{57}{639}=0.0892$ represent the proportion of correct answers for university non-degree holders

z would represent the statistic (variable of interest)

$p_v$ represent the value for the test (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportions are different between the two groups, the system of hypothesis would be:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{45+57}{373+639}=0.101$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.121-0.0892}{\sqrt{0.101(1-0.101)(\frac{1}{373}+\frac{1}{639})}}=1.620$

Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test.

Since is a one side test the p value would be:

$p_v =2*P(Z>1.620)=0.105$

7 0

2 years ago

PLEASE HELP! ASAP! URGENT!

grandymaker [24]

Answer:

g(-7)= -37

Step-by-step explanation:

g(y)= 7y + 12

plug in -7

g(-7)= 7(-7) + 12

simplify

g(-7)= -49 +12

g(-7)= -37

6 0

2 years ago

You deposit $300 in a savings account that pays 6% interest compounded semiannually. How much will you have at the middle of the

Otrada [13]

Answer:

Please check the explanation.

Step-by-step explanation:

a) How much will you have at the middle of the first year?

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 0.5 years

Total amount = A = ?

Using the formula

$A\:=\:P\left(1+\frac{r}{n}\right)^{nt}$

substituting the values

$A=300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(0.5\right)}$

$A=300\cdot \frac{2.06}{2}$

$A=\frac{618}{2}$

$A=309$ $

Therefore, the total amount accrued, principal plus interest, from compound interest on an original principal of $ 300.00 at a rate of 6% per year compounded 2 times per year over 0.5 years is $ 309.00.

Part b) How much at the end of one year?

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 1 years

Total amount = A = ?

Using the formula

$A\:=\:P\left(1+\frac{r}{n}\right)^{nt}$

so substituting the values

$A\:=\:300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(1\right)}$

$A=300\cdot \frac{2.06^2}{2^2}$

$A=318.27$ $

Therefore, the total amount accrued, principal plus interest, from compound interest on an original principal of $ 300.00 at a rate of 6% per year compounded 2 times per year over 1 year is $ 318.27.

7 0

2 years ago

HELP PLS<br> Is this set of ordered pairs a function <br> (8,4) (6,2) (1,9) (3,8) (0,7)

Mrrafil [7]

Answer:

Step-by-step explanation:

They are not being added by the same number each time.

4 0

3 years ago

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