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2 years ago

Can someone answer this question please answer it correctly if it’s corect I will mark you brainliest

Mathematics

2 answers:

drek231 [11]2 years ago

5 0

Answer:

All of the above

Step-by-step explanation:

There is one Queen of hearts in a 52-card deck

Write as a fraction

1/52

Write as a decimal

~0.019

Convert to a percentage (multiply by 100)

The three (equivalent) forms used to represent to probability:

1/52, 0.019, 2%

These are all of the answer choices provided, so the answer would be all of the above

Hope this helps :)

Firdavs [7]2 years ago

5 0

Answer:

All of the above

Step-by-step explanation:

There is 1 queen of hearts in a deck of 52 cards

P (queen of hears) = 1/ 52 fraction

=.019 decimal

1.9% percent

All of the above

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Kadin and Silas went to a movie that started at 1:55 p.m. and got out at 4:10 pm

elixir [45]

The movie should be 2 hours and 15 minutes long.

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2 years ago

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Suppose 2 /7 of students in your school wore shorts today. How many students would you have to select randomly, on average, in o

Sveta_85 [38]

Answer: 49 students

Step-by-step explanation:

2 /7 of students in a school wore shorts today. The number of students to be selected in order to get 14 students who will wear shorts to school will be:

Let the number of students that will be sampled be y. This will be:

2/7 of y = 14

2/7 × y = 14

2y/7 = 14

Cross Multiply

2y = 7 × 14

2y = 98

y = 98/2

y = 49

49 students will be selected randomly.

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3 years ago

Help! Prove the equality arccos √(2/3) - arccos (1+√6)/(2*√3) = π/6

stealth61 [152]

Answer:

Proof in the explanation

Step-by-step explanation:

Trigonometric Equalities

Those are expressions involving trigonometric functions which must be proven, generally working on only one side of the equality

For this particular equality, we'll use the following equation

$\displaystyle cos(x-y)=cos\ x\ cos\ y+sin\ x\ sin\ y$

The equality we want to prove is

$\displaystyle arccos\ \sqrt{\frac{2}{3}}-arccos\left(\frac{1+\sqrt{6}}{2\sqrt{3}}\right)=\frac{\pi}{6}$

Let's set the following variables:

$\displaystyle x=arccos\ \sqrt{\frac{2}{3}},\ y=arccos(\frac{1+\sqrt{6}}{2\sqrt{3}})$

And modify the first variable:

$\displaystyle x=arccos\ \frac{\sqrt{6}}{3}}=>\ cos\ x= \frac{\sqrt{6}}{3}}$

Now with the second variable

$\displaystyle y=arccos\ \frac{1+\sqrt{6}}{2\sqrt{3}}=>cos\ y=\frac{1+\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{3}+3\sqrt{2}}{6}$

Knowing that

$sin^2x+cos^2x=1$

We compute the other two trigonometric functions of X and Y

$\displaystyle sin \ x=\sqrt{1-cos^2\ x}=\sqrt{1-(\frac{\sqrt{6}}{3})^2}=\sqrt{1-\frac{6}{9}}=\frac{\sqrt{3}}{3}$

$\displaystyle sin\ y=\sqrt{1-cos^2y}=\sqrt{1-\frac{(\sqrt{3}+3\sqrt{2})^2}{36}}}$

$\displaystyle sin\ y=\sqrt{\frac{36-(3+6\sqrt{6}+18)}{36}}=\sqrt{\frac{15-6\sqrt{6}}{36}}$

Computing

$15-6\sqrt{6}=(3-\sqrt{6})^2$

Then

$\displaystyle sin\ y=\frac{3-\sqrt{6}}{6}$

Now we replace all in the first equality:

$\displaystyle cos(x-y)=\frac{\sqrt{6}}{3}.\frac{\sqrt{3}+3\sqrt{2}}{6}+\frac{\sqrt{3}}{3}.\frac{3-\sqrt{6}}{6}$

$\displaystyle cos(x-y)=\frac{3\sqrt{2}+6\sqrt{3}}{18}+\frac{3\sqrt{3}-3\sqrt{2}}{18}$

$\displaystyle cos(x-y)=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}=cos\ \pi/6$

Thus, proven

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3 years ago

X + 2 2/1=10 HELPPPPPPP!!!!!!!!!!!!!!!!!

irakobra [83]

Answer:

x+4=10
x=6
Give brainlist

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2 years ago

A student answered three questions on a 40-question test incorrectly. What percent did the student answer correctly?

Sveta_85 [38]

the numerator is 37; the denominator is 40

37/40=.925

move the decimal two spots to the right and you get 92.5%

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2 years ago

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