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3 years ago

Pls answer this for meee

Mathematics

2 answers:

klasskru [66]3 years ago

7 0

Answer:

Step-by-step explanation:

its perpedicular so it adds up to 90°, therefore 90-55=x and x =35

goldfiish [28.3K]3 years ago

5 0

Answer:

x=35 degrees

Step-by-step explanation:

Since the other side is 90 degrees you would have to add 55 degrees + 35 degrees in order to get 90 degrees for both sides.

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Solve the equation: 13W - 2(4W + 1) = W - 58 *

Sergio [31]

Answer:

W = - 14

Step-by-step explanation:

13W - 2(4W + 1) = W - 58

13W - 8W - 2 = W - 58

13W - 8W - W = 2 - 58

4W = - 56

W = - 56 : 4

W = - 14

6 0

3 years ago

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Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

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4 years ago

ALGEBRA 2 HOW DO YOU LIST THE TRANSFORMATIONS OF THE GRAPH GIVEN THE EQUATION??

Fed [463]

Answer:

Step-by-step explanation:

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3 years ago

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Find the area of a circle with a diameter of 6.

Kazeer [188]

Answer:

A = 9pi

Step-by-step explanation:

The area of a circle is given by

A = pi r^2

We know the diameter

r = d/2 = 6/2 = 3

A = pi (3)^2

A = 9pi

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4 years ago

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What do I type in the yellow boxes?

MA_775_DIABLO [31]

Answer:

the first yellow box would be -14

the second one would be 2

the last one would be -7!

hope this helped!

Step-by-step explanation:

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2 years ago

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