A bag contains 8 blue coins and 7 red coins A coin is removed at random what is the probability that the coin removed is blue

ASHA 777 [7]

5y + 3 > -7 + 13

First add -7 + 13 = 6

5y + 3 > 6

Subtract 3 from both sides

5y + 3 -3 > 6 - 3

5y > 3

Divide both sides by 5

5y/5 > 3/5

y > 3/5

5 0

2 years ago

What value of x is in the solution set of 3(x – 4) ≥ 5x 2?

Andrej [43]

Hope there are

3(x-4) >= 5x^2 or this 2 from the end what mean ?

3x-12 >=5x^2

0>=5xˇ2 -3x +12

5x^2 -3x +12 =0

D = 9 -240 = -150 so what mean that this equation has not real solutions

5 0

2 years ago

Read 2 more answers

You choose a marble at random from a bag containing 8 brown marbles 6 yellow marbles and one purple marble. You replace the marb

Andrej [43]

Pick 1: 6 yellow out of 15 total marbles = 6/15 = 2/5
Pick 2: 6 yellow out of 15 total marbles = 6/15 = 2/5

Pick 1 AND Pick 2
2/5 x 2/5 = 4/25

5 0

2 years ago

The probability that a pupil in class has a calculator is 0.87 what is the probability that a pupil does not have a calculator?

lidiya [134]

Answer:

0.87 1.0

0.13

Step-by-step explanation:

0.13 dont have a calculator

8 0

2 years ago

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .

Equate the right-hand side of these two equations:

$0.810\, x \approx 0.649\, (x + 20)$ .

Solve for $x$ :

$x \approx 81\; \rm ft$ .

Hence, the height of the top of this tree relative to the base of the hill would be $(x + 20)\; {\rm ft}\approx 101\; \rm ft$ .

6 0

2 years ago