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3 years ago

Anyone know this?????

Mathematics

1 answer:

kifflom [539]3 years ago

6 0

Answer:

WZX=57

Step-by-step explanation:

90-33 = 57

You don't have to use the equation unless you want to check yourself.

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each classroom in willett hall will hold 34 students.There are 4 floors with 6 classroom on each floor. How many students can be

ozzi

4 floors with 6 class rooms each = 6*4 = 24 total class rooms.

24 class rooms x 34 students in each one = 816 total students.

4 0

3 years ago

Read 2 more answers

From a height of 50 meters above sea level on a cliff, two ships are sighted due west. The angles of depression are 61° and 28°.

Olenka [21]

Answer:

The ships are 66 meters apart.

Step-by-step explanation:

For the sake of convenience, let us label ships A and B

As shown in the figure, the distances to the ships from right triangles.

The distance to the ship A is $d_1$ and it is given by

$tan (61^o)= \dfrac{50}{d_1}$

$d_1=\dfrac{50}{tan (61^o)}$

$\boxed{d_1= 27.71m}$

And the distance to the ship B is $d_2$ and is given by

$tan (28^o)= \dfrac{50}{d_2}$

$d_2=\dfrac{50}{tan (28^o)}$

$\boxed{ d_2=94.04m}$

Therefore, the distance $d$ between the ships A and B is

$d= d_2-d_1=94.04-27.7\\\\\boxed{d=66m}$

In other words, the ships are 66 meters apart.

8 0

3 years ago

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During which one-day interval did she work the most hours?

Karo-lina-s [1.5K]

Um well lets see there is no question so the day that ends in y

5 0

3 years ago

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How do I solve for the quadratic function?

Basile [38]

Aw you put the numbers in so a b c then you solve

3 0

3 years ago

If a coin is tossed three times, find probability of getting

Assoli18 [71]

${\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}$

‣ A coin is tossed three times.

${\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}$

‣ The probability of getting,

1) Exactly 3 tails

2) At most 2 heads

3) At least 2 tails

4) Exactly 2 heads

5) Exactly 3 heads

${\large{\textsf{\textbf{\underline{\underline{Using \: Formula :}}}}}}$

$\star \: \tt P(E)= {\underline{\boxed{\sf{\red{ \dfrac{ Favourable \: outcomes }{Total \: outcomes} }}}}}$

${\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}$

★ When three coins are tossed,

then the sample space = {HHH, HHT, THH, TTH, HTH, HTT, THT, TTT}

[here H denotes head and T denotes tail]

⇒Total number of outcomes $\tt [ \: n(s) \: ]$ = 8

1) Exactly 3 tails 

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

$\therefore \sf Probability_{(exactly \: 3 \: tails)} = \red{ \dfrac{1}{8}}$

2) At most 2 heads

[It means there can be two or one or no heads]

Here

• Favourable outcomes = {HHT, THH, HTH, TTH, HTT, THT, TTT} = 7

• Total outcomes = 8

$\therefore \sf Probability_{(at \: most \: 2 \: heads)} = \green{ \dfrac{7}{8}}$

3) At least 2 tails 

[It means there can be two or more tails]

Here

• Favourable outcomes = {TTH, TTT, HTT, THT} = 4

• Total outcomes = 8

$\longrightarrow \sf Probability_{(at \: least \: 2 \: tails)} = \dfrac{4}{8}$

$\therefore \sf Probability_{(at \: least \: 2 \: tails)} = \orange{\dfrac{1}{2}}$

4) Exactly 2 heads 

Here

• Favourable outcomes = {HTH, THH, HHT } = 3

• Total outcomes = 8

$\therefore \sf Probability_{(exactly \: 2 \: heads)} = \pink{ \dfrac{3}{8}}$

5) Exactly 3 heads

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

$\therefore \sf Probability_{(exactly \: 3 \: heads)} = \purple{ \dfrac{1}{8}}$

$\rule{280pt}{2pt}$

8 0

1 year ago