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1 year ago

Help? Use the law of sines to find the length of side c

Mathematics

2 answers:

____ [38]1 year ago

8 0

Answer:

Step-by-step explanation:

Using the law of Sines in ΔABC

$\frac{a}{sinA}$ = $\frac{c}{sinC}$ , that is

$\frac{37}{sin42}$ = $\frac{c}{sin41.5}$ ( cross- multiply )

c × sin42° = 37 × sin41.5° ( divide both sides by sin42° )

c = $\frac{37(sin41.5)}{sin42}$ ≈ 36.64 ( to 2 dec. places )

Ahat [919]1 year ago

5 0

<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ a/sinA = c/sinC

Substitute in the values:

37/sin(42) = c/sin(41.5)

Multiply both sides by sin(41.5)

37/sin(42) x sin(41.5) = c

Solve:

c = 36.63999457

The correct answer would be C. 36.64

➶ Hope This Helps You!

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Pediatric asthma survey, n = 50. suppose that asthma affects 1 in 20 children in a population. you take an srs of 50 children fr

zaharov [31]

In this situation,
n=50, p=1/20, q=(1-p)=19/20, and npq=19/8=2.4

We would like np and npq to be a large number, at least greater than 10.
The normal approximation can always be applied, but the result will be very approximate, depending on the values of np and npq.

Situations are favourable for the normal approximation when p is around 0.5, say between 0.3 and 0.7, and n>30.

"Normal approximation" is using normal probability distribution to approximate the binomial distribution, when n is large (greater than 70) or exceeds the capacity of most hand-held calculators. The binomial distribution can be used if the following conditions are met:
1. Bernoulli trials, i.e. exactly two possible outcomes.2. Number of trials is known before and constant throughout the experiment, i.e. independent of outcomes.3. All trials are independent of each other.4. Probability of success is known, and remain constant throughout trials.
If all criteria are satisfied, we can model with binomial distribution, where the probability of x successes out of N trials each with probability of success p is given byP(x)=C(N,x)(p^x)(1-p)^(N-x)and,C(N,x) is number of combinations of selecting x objects out of N.
The mean is np, and variance is npq.

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1 year ago

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1 year ago

A triangle has a base length of 3.1 cm and corresponding height of 4.2 cm, correct to

Svetradugi [14.3K]

Answer:

The upper bound of the area is 6.69375 cm²

The lower bound of the area is 6.32875 cm²

Step-by-step explanation:

The question relates to measurements and accuracy

The given information are;

The base length of the triangle = 3.1 cm

The height of the triangle = 4.2 cm correct to one decimal place

The lower bound for the base length of the triangle = 3.05 cm

The upper bound for the base length of the triangle = 3.15 cm

Similarly, the lower bound for the height of the triangle = 4.15 cm

The upper bound for the height of the triangle = 4.25 cm

The formula for calculating the area of a triangle is given by the following formula;

Area of a triangle = 1/2 × Base × Height

The upper bound of the area = 1/2 × The upper bound of the base × The upper bound of the height

The upper bound of the area = 1/2 × 3.15 × 4.25 = 6.69375 cm²

The upper bound of the area = 6.69375 cm²

The lower bound of the area = 1/2 × The lower bound of the base × The lower bound of the height

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2 years ago

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yanalaym [24]

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1 year ago

A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the pr

Gre4nikov [31]

Answer:

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$z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5$

And using a calculator, excel ir the normal standard table we have that:

$P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)$

And we can calculate the probability like this: $P(-1.5 \leq Z \leq 1.5) = P(z$ Step-by-step explanation:

A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval 47<=X<53. Is the assumption of normality important. Why?

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(50,12)$