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3 years ago

Please answer !!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!

Mathematics

1 answer:

Tom [10]3 years ago

4 0

Answer:

x = 111 degrees

Step-by-step explanation:

the angle next to 164 degrees :

any line = 180 degrees, so 180 - 164 = 16 degrees

now you have two out of three angles in the triangle. All three angles of every triangle must add together to equal 180, so

180 = 53 + 16 + x

x= 180 - 53 - 16

x=111 degrees

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The function f(x)=2x^2-8x+7 has a maximum of how many turns?

Gnesinka [82]

Answer:

Step-by-step explanation:

This is a quadratic function which is parabolic in shape. By definition, it has a vertex which serves as the (only) max or min point; 1 turn.

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4 years ago

Please can u help me with geometry work !!!

alisha [4.7K]

The first one can be done by Pythagoras theorem such as $12^2=9^2+x^2$ then $x^2=63$ then $x= \sqrt{63}$

The second one can also done by Pythagoras theorem, note that this is isosceles triangle so $y=z$ , again by using Pythagoras theorem $15^2=y^2+z^2=y^2+y^2 or =z^2+z^2$ then $225=2y^2$ then $y= \sqrt{\frac{225}{2} }$

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3 years ago

What to say back if someone says that you’re skinny

MissTica

Answer:

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Step-by-step explanation:

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3 years ago

Consider the following differential equation. (A computer algebra system is recommended.) (1 + t2)y' + 4ty = (1 + t2)−2 (a) Draw

alina1380 [7]

Answer:

Step-by-step explanation:

a. Draw a direction field for the given differential equation

b. Based on the inspection of the direction field, describe ow solutions behave for large t.

The solution appear oscillatory

All solutions seems to converge to the function y0(t)=4

All solutions seems to converge to the function y0(t)=0

All solutions seems to seems to eventually have negative slopes a and hence decrease without bound

All solutions seems to seems to eventually have positive slopes a and hence increase without bound

As t-infinity

All solutions seems to seems to eventually have positive slopes a and hence decrease without bound

All solutions seems to converge to the function y0(t)=0

All solutions seems to seems to eventually have negative slopes a and hence decrease without bound

All solutions seems to converge to the function y0(t)=4

The solution are oscillatory

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3 years ago

At NC State University, 16.4% of the undergraduate classes have more than 50 students. If a random sample of 200 undergraduate c

Lena [83]

Answer:

We need to check the conditions in order to use the normal approximation.

$np=200*0.164=32.8 \geq 10$

$n(1-p)=200*(1-0.164)=167.2 \geq 10$

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

The mean is given by:

$p =0.164$

And the deviation is given by:

$\sigma_{p}= \sqrt{\frac{0.164*(1-0.164)}{200}}= 0.0262$

So then the correct options for this case are:

The sampling distribution will be approximately normal.

The mean of the sampling distribution will be 16.4%.

The standard deviation of the sampling distribution will be 0.0262.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=200, p=0.164)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

We need to check the conditions in order to use the normal approximation.

$np=200*0.164=32.8 \geq 10$

$n(1-p)=200*(1-0.164)=167.2 \geq 10$

The population proportion have the following distribution

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

The mean is given by:

$p =0.164$

And the deviation is given by:

$\sigma_{p}= \sqrt{\frac{0.164*(1-0.164)}{200}}= 0.0262$

So then the correct options for this case are:

The sampling distribution will be approximately normal.

The mean of the sampling distribution will be 16.4%.

The standard deviation of the sampling distribution will be 0.0262.

6 0

3 years ago