Hi<br><br>answer correct and I will follow you

What is the value of the 4 in 3.954

antiseptic1488 [7]

Answer:

The value is ones lol thats easy mark me brainliest plzzz

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Which equation relates the number of triangles in the figure (n) to the perimeter of the figure (p)

WITCHER [35]

Answer:

Option d is correct.

Equation : P=7n+20.

Explanation;

Given the perimeter of each figure is;

Perimeter of triangle is equal to the sum of all the sides of a triangle.

Perimeter of 1 triangle is 21

Perimeter of 2 triangle is 34 and

Perimeter of 3 triangle is 41

Let n be the number of figure and P be the perimeter of the figure;

the only equation which satisfy the given perimeter is;

$P=7n+20$

Check:

for n =1 which means 1 triangle then;

$P =7n+20 = 7 \cdot 1 +20 = 7+20 = 27$

for n = 2 , [ i.e 2 triangle]

$P =7n+20 = 7 \cdot 2 +20 = 14+20 = 34$

and for n =3 [i.e, 3 triangles]

$P =7n+20 = 7 \cdot 3 +20 = 21+20 = 41$

Therefore, the equation P =7n+20 relates the number of triangles in the figure(n) to the perimeter of the figure(P).

3 0

3 years ago

If the probability of a student taking a calculus class is 0.10, the probability of taking a statistics class is 0.90, and the p

Stells [14]

Answer:

93% probability of a student taking a calculus class or a statistics class

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student takes a calculus class.

B is the probability that a student takes a statistics class.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student takes calculus but not statistics and $A \cap B$ is the probability that a student takes both these classes.

By the same logic, we have that:

$B = b + (A \cap B)$

The probability of taking a calculus class and a statistics class is 0.07

This means that $A \cap B = 0.07$

The probability of taking a statistics class is 0.90

This means that $B = 0.9$ . So

$B = b + (A \cap B)$

$0.9 = b + 0.07$

$b = 0.83$

The probability of a student taking a calculus class is 0.10

This means that $A = 0.1$

$A = a + (A \cap B)$

$0.1 = a + 0.07$

$a = 0.03$

What is the probability of a student taking a calculus class or a statistics class

$A \cup B = a + b + A \cap B = 0.03 + 0.83 + 0.07 = 0.93$

93% probability of a student taking a calculus class or a statistics class

5 0

3 years ago

You are scheduled to receive $38,000 in two yeara. when you receive it, you will invest it for 10 more years at 6.0 percent per

Savatey [412]

You will have 38,0000 dollars

7 0

3 years ago

Explorers on a planet with a thin atmosphere want to measure the acceleration of gravity at its surface. So they use a spring gu

Murljashka [212]

Answer:

The value of $g_s$ is $g_s=\frac{30}{21} \:\frac{m}{s^2}$ .

Step-by-step explanation:

We know that the position function is given by

$s(t)=30t-\frac{g_s}{2} t^2$

Velocity is defined as the rate of change of position. Therefore,

$v(t)=\frac{d}{dt}(s(t))$

So, the velocity function is

$v(t)=\frac{d}{dt}(s(t))\\\\v(t)=\frac{d}{dt}(30t-\frac{g_s}{2} t^2)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(30t\right)-\frac{d}{dt}\left(\frac{g_s}{2}t^2\right)\\\\v(t)=30-g_st$

When a body reaches a vertical velocity of zero, this is the maximum height of the body and then gravity will take over and accelerate the object downward. Thus,

$v(t)=30-g_st=0\\\\g_s=\frac{30}{t}$

We know that the ball bearing reaches its maximum height 21 seconds after being launched (t = 21 s).

So, the value of $g_s$ is $g_s=\frac{30}{21} \:\frac{m}{s^2}$ .

5 0

3 years ago

Hianswer correct and I will follow you​

Hianswer correct and I will follow you