Can someone please help me?

On her way to work, Mrs. Larsen gets a green light at a certain traffic light only 30% of the time. She wants to find the probab

vladimir1956 [14]

First, use the random number table to generate the possible probabilities but only account the first four digits in order to get a value that is a percentage
35.85%
28.97%
90.12%
5.07%
20.43%
The average is
36.09%
There is a 36.09% probability that the light will be green when reaches the traffic light

7 0

2 years ago

25x² - 4 <br>I'm trying to help a friend n that's all they gave me ahh

torisob [31]

This is a difference of squares problem. You just have to make sure every term is a perfect square and that there is subtraction in the problem.

7 0

3 years ago

There is a paint chip on the edge of a gear for a music box. It rotates to (−7, 3) during one song when represented graphically.

lys-0071 [83]

I hope this helps you

6 0

3 years ago

A rectangle has an area of 19.38 cm2. When both the length and width of the rectangle are increased by 1.50 cm, the area of the

Lostsunrise [7]

Answer: $5.7\ cm$

Step-by-step explanation:

Given

Rectangle has an area of $19.38\ cm^2$

Suppose rectangle length and width are $l$ and $w$

If each side is increased by $1.50\ cm$

Area becomes $A_2=35.28\ cm^2$

We can write

$\Rightarrow lw=19.38\quad \ldots(i)\\\\\Rightarrow (l+1.5)(w+1.5)=35.28\\\Rightarrow lw+1.5(l+w)+1.5^2=35.28\\\text{use (i) for}\ lw\\\Rightarrow 19.38+1.5(l+w)=35.28-2.25\\\Rightarrow l+w=9.1\quad \ldots(ii)$

Substitute the value of width from (ii) in equation (i)

$\Rightarrow l(9.1-l)=19.38\\\Rightarrow l^2-9.1l+19.38=0\\\\\Rightarrow l=\dfrac{9.1\pm\sqrt{(-9.1)^2-4(1)(19.38)}}{2\times 1}\\\\\Rightarrow l=\dfrac{9.1\pm\sqrt{5.29}}{2}\\\\\Rightarrow l=\dfrac{9.1\pm2.3}{2}\\\\\Rightarrow l=3.4,\ 5.7$

Width corresponding to these lengths

$w=5.7,\ 3.4$

Therfore, we can write the length of the longer side is $5.7\ cm$

8 0

3 years ago

[6+4=10 points] Problem 2. Suppose that there are k people in a party with the following PMF: • k = 5 with probability 1 4 • k =

kirza4 [7]

Answer:

1). 0.903547

2). 0.275617

Step-by-step explanation:

It is given :

K people in a party with the following :

i). k = 5 with the probability of $\frac{1}{4}$

ii). k = 10 with the probability of $\frac{1}{4}$

iii). k = 10 with the probability $\frac{1}{2}$

So the probability of at least two person out of the 'n' born people in same month is = 1 - P (none of the n born in the same month)

= 1 - P (choosing the n different months out of 365 days) = $1-\frac{_{n}^{12}\textrm{P}}{12^2}$

1). Hence P(at least 2 born in the same month)=P(k=5 and at least 2 born in the same month)+P(k=10 and at least 2 born in the same month)+P(k=15 and at least 2 born in the same month)

= $\frac{1}{4}\times (1-\frac{_{5}^{12}\textrm{P}}{12^5})+\frac{1}{4}\times (1-\frac{_{10}^{12}\textrm{P}}{12^{10}})+\frac{1}{2}\times (1-\frac{_{15}^{12}\textrm{P}}{12^{15}})$

= $0.25 \times 0.618056 + 0.25 \times 0.996132 + 0.5 \times 1$

= 0.903547

2).P( k = 10|at least 2 share their birthday in same month)

=P(k=10 and at least 2 born in the same month)/P(at least 2 share their birthday in same month)

= $0.25 \times \frac{0.996132}{0.903547}$

= 0.0.275617

6 0

3 years ago