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1 year ago

Can someone help me with this question? Thank you!

Mathematics

2 answers:

Harman [31]1 year ago

6 0

Answer:

122°

Step-by-step explanation:

The Angle Sum Property of a triangle gives that the sum of the internal angles of a triangle is equal to 180°.

Hence,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + <u>43°</u> + <u>15°</u> = 180°

⇒ ∠A + 58° = 180°

⇒ ∠A = 122°

emmasim [6.3K]1 year ago

4 0

Answer:

122°

Step-by-step explanation:

triangle = 180°

∴180 - 43 - 15

=122°

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A 13 Ft ladder is leaning up against a building, with the base of the ladder 2 feet from the building. How high up on the buildi

Lelechka [254]

Visualize the image in your mind and you can see that it forms a right angled triangle with a base,altitude and hypotenuse

base=2 feet

hypotenuse(ladder)=13 ft

We need to find the altitude,

By applying pythagoras theorem,

base square+altitude square=hypotenuse square

Take altitude as x

2 square+x square=13 square

4+x square=169

x square=169-4

x square=165

x= root of 165

x=12.8 feet(approx.)

6 0

3 years ago

Information about the proportion of a sample that agrees with a certain statement is given below. Use StatKey or other technolog

lara31 [8.8K]

Answer:

$SE=\sqrt{\frac{\hat p (1-\hat p)}{n}}=\sqrt{\frac{0.35 (1-0.35)}{100}}=0.048$

If we replace the values obtained we got:

$0.35 - 1.96\sqrt{\frac{0.35(1-0.35)}{100}}=0.257$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.443$

The 95% confidence interval would be given by (0.257;0.443)

Step-by-step explanation:

Notation and definitions

$X=35$ number of people that agree.

$n=100$ random sample taken

$\hat p=\frac{35}{100}=0.35$ estimated proportion of people that agree

$p$ true population proportion of peopl that agree

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: