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Trava [24]
3 years ago
10

Please please help meeee

Mathematics
2 answers:
Sladkaya [172]3 years ago
8 0

Answer:

16

Step-by-step explanation:

10/a = 15/24

Cross multiply.

15 × a = 10 × 24

15a = 240

Divide both sides with 15.

a = 240/15

a = 16

larisa [96]3 years ago
3 0

Answer:

a = 16

Step-by-step explanation:

=> \frac{10}{a} = \frac{15}{24}

Cross Multiplying

=> 10 * 24 = a * 15

=> 15a = 240

Dividing both sides by 15

=> a = 16

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Let the random variable x be equally likely to assume any of the values 1/10, 1/5, or 3/10. determine the mean and variance of x
Keith_Richards [23]
Since the values are equally likely and equally spaced, their mean is the middle one.
  the mean is 1/5

The variance is the average of the square of the deviations from the mean.
  σ² = (1/3)((-1/10)² + 0² + (1/10)²) = 2/300
  the variance is 2/300
6 0
3 years ago
What is the measure of ∠ADC in quadrilateral ABCD? °
anastassius [24]
115 is the correct answer.
I saw the diagram earlier :)
Hope this helps, have a great day, and God bless.
Brainliest is always appreciated :)
7 0
4 years ago
looks out from the crown of the Statue of Liberty, approximately 250 ft above ground. The tourist sees a ship coming into the ha
Zigmanuir [339]

Answer:

The distance from the base of the statue to the ship is 771.60 ft

Step-by-step explanation:

Refer the attached figure

Height of statue AB= 250 feet

The tourist sees a ship coming into the harbor and measures the angle of depression as 18°.

So, ∠ACB = 18°

We are supposed to find the distance from the base of the statue to the ship i.e. BC

In ΔABC

Tan \theta = \frac{Perpendicular}{Base}

Tan 18^{\circ}=\frac{AB}{BC}

Tan 18^{\circ}=\frac{250}{BC}

BC=\frac{250}{Tan 18^{\circ}}

BC=\frac{250}{0.324}

BC=771.60 ft

Hence the distance from the base of the statue to the ship is 771.60 ft

3 0
3 years ago
What is (3x+5)(x^2+6x+11)
serg [7]

Answer:

\large\boxed{3x^3+23x^2+63x+55}

Step-by-step explanation:

\text{Use FOIL:}\\\\(a+b)(c+d)=ac+ad+bc+bd

(3x+5)(x^2+6x+11)\\\\=(3x)(x^2)+(3x)(6x)+(3x)(11)+(5)(x^2)+(5)(6x)+(5)(11)\\\\=3x^3+18x^2+33x+5x^2+30x+55\qquad\text{combine like terms}\\\\=3x^3+(18x^2+5x^3)+(33x+30x)+55\\\\=3x^3+23x^2+63x+55

3 0
4 years ago
I need helppppppp. whats 1/5x30
Vadim26 [7]

Answer:

6

Step-by-step

You're welcome!

:)

3 0
3 years ago
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