What is 45% written as a fraction

Given the information in the diagram below, find the measure of angle N. Use RACE and complete sentences to show how you found t

Sonja [21]

Answer:

m<N = 76°

Step-by-step explanation:

Given:

∆JKL and ∆MNL are isosceles ∆ (isosceles ∆ has 2 equal sides).

m<J = 64° (given)

Required:

m<N

SOLUTION:

m<K = m<J (base angles of an isosceles ∆ are equal)

m<K = 64° (Substitution)

m<K + m<J + m<JLK = 180° (sum of ∆)

64° + 64° + m<JLK = 180° (substitution)

128° + m<JLK = 180°

subtract 128 from each side

m<JLK = 180° - 128°

m<JLK = 52°

In isosceles ∆MNL, m<MLN and <M are base angles of the ∆. Therefore, they are of equal measure.

Thus:

m<MLN = m<JKL (vertical angles are congruent)

m<MLN = 52°

m<M = m<MLN (base angles of isosceles ∆MNL)

m<M = 52° (substitution)

m<N + m<M° + m<MLN = 180° (Sum of ∆)

m<N + 52° + 52° = 180° (Substitution)

m<N + 104° = 180°

subtract 104 from each side

m<N = 180° - 104°

m<N = 76°

4 0

2 years ago

Bruh, who else has exams?

Llana [10]

Answer:

I have exams too well cba's

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

An engineer on the ground is looking at the top of a building. The angle of elevation to the top of a buliding. The angle of ele

Marysya12 [62]

$\tan( \alpha ) \: = \: \frac{h}{b}$
$\alpha \: = \: {22}^{0}$
$h \: = \: 450 \: ft$
Therefore,
$b \: = \: h \cot( \alpha )$
$b \: = \: 450 \: ft \: \times \: \cot( {22}^{0} )$
b = 450 ft × 2.47508685
b = 1,113.78908 ft
or,
b ≈ 1,114 ft.

6 0

2 years ago

If a survey of a movie shows that 50% of all people who like the movie, enjoyed the official movie soundtrack too. If a total of

Murljashka [212]

Answer:

For the first question the sample size is n= 3850

$ME= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$\hat p \pm ME$

And replacing we got:

$0.5- 0.016 = 0.484$

$0.5+ 0.016 = 0.516$

We are confident at 95% that the true proportion of all people who like the movie is between 0.484 and 0.516.

Step-by-step explanation:

We have the following info given:

$\hat p = 0.5$ represent the estimated proportion of all people who like the movie

$n = 3850$ represent the sample size selected

For the first question the sample size is n= 3850

For the second part they gives to us the margin of error:

$ME = 0.016$

We know that the confidence interval for a population proportion $p$ of interest is given by:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

And the margin of error is given by:

$ME= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

So then the confidence interval can be found with this formula:

$\hat p \pm ME$

And replacing we got:

$0.5- 0.016 = 0.484$

$0.5+ 0.016 = 0.516$

We are confident at 95% that the true proportion of all people who like the movie is between 0.484 and 0.516.

5 0

2 years ago

HELP!!

tamaranim1 [39]

Question 1: The answer is D. which it ended up being 0.9979

Question 2: The expression P(z > -0.87) represents the area under the standard normal curve above a given value of z. What is P(z > -0.87)? Express your answer as a decimal to the nearest ten thousand

The expression P(z > -0.87) represents the area under the standard normal curve above a given value of z. What is P(z > -0.87)? Express your answer as a decimal to the nearest ten thousandth (four decimal places). So being that rounding it off would mean your answer would be = ?

Question 3: Assume that the test scores from a college admissions test are normally distributed, with a mean of 450 and a standard deviation of 100.
a. What percentage of the people taking the test score between 400 and 500?
b. Suppose someone receives a score of 630. What percentage of the people taking the test score better? What percentage score worse?
c. A university will not admit a student who does not score in the upper 25% of those taking the test regardless of other criteria. What score is necessary to be considered for admission? z = 600-450 /100 = .5 NORMSDIST(0.5) = .691462 z = 400-450 /100 = -.5 NORMSDIST(-0.5) = .30854

P( -.5 < z <.5) = .691462 - .30854 = .3829 Or 38.29%
Receiving score of 630:
z = 630-450 /100 = 1.8 NORMSDIST(1.8) = .9641
96.41% score less and 3.59 % score better
upper 25%
z = NORMSINV(0.75)= .6745
.6745 *100 + 450 = 517 Would need score >517 to be considered for admissions 
Question 4: The z-score for 45cm is found as follows:
Reference to a normal distribution table, gives the cumulative probability as 0.0099.
Therefore about 1% of newborn girls will be 45cm or shorter.

6 0

3 years ago

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