Answer:
In x - 5 years.
Step-by-step explanation:
5 years ago:
Sam's age: x
Nina's age: 2x
Now:
Sam's age: x + 5
Nina's age: 2x + 5
Let t be the number of years from now that Nina will be 1.5 times Sam's age.
In t years:
Sam's age: x + t + 5
Nina's age: 2x + t + 5
Nina will be 1.5 times Sam's age:
2x + t + 5 = 1.5(x + t + 5)
2x + t + 5 = (3/2)(x + t + 5)
2(2x + t + 5) = 3(x + t + 5)
4x + 2t + 10 = 3x + 3t + 15
4x + 10 = 3x + t + 15
x - 5 = t
t = x - 5
Answer:

Step-by-step explanation:
Consider the equation

Substitute

into the equation:

This is quadratic equation, where

Hence,

Now, use the substitution again:
1. When
then

2. When
then

This equation has no solutions, because the range of
is
and 3>1.
9.8 m/s
I'm not sure but I think this is the correct answer
Answer:
The lower bound is,
and the upper bound is
.
Step-by-step explanation:
Let the random variable <em>X</em> follows a normal distribution with mean <em>μ </em>and standard deviation <em>σ</em>.
The the random variable <em>Z, </em>defined as
is standardized random variable also known as a standard normal random variable. The random variable
.
The standard normal random variable has a symmetric distribution.
It is provided that
.
Determine the upper and lower bound as follows:
![P(-z\leq Z\leq z)=0.51\\P(Z\leq z)-P(Z\leq -z)=0.51\\P(Z\leq z)-[1-P(Z\leq z)]=0.51\\2P(Z\leq z)-1=0.51\\2P(Z\leq z)=1.51\\P(Z\leq z)=0.755](https://tex.z-dn.net/?f=P%28-z%5Cleq%20Z%5Cleq%20z%29%3D0.51%5C%5CP%28Z%5Cleq%20z%29-P%28Z%5Cleq%20-z%29%3D0.51%5C%5CP%28Z%5Cleq%20z%29-%5B1-P%28Z%5Cleq%20z%29%5D%3D0.51%5C%5C2P%28Z%5Cleq%20z%29-1%3D0.51%5C%5C2P%28Z%5Cleq%20z%29%3D1.51%5C%5CP%28Z%5Cleq%20z%29%3D0.755)
Use a standard normal table to determine the value of <em>z.</em>
The value of <em>z</em> such that P (Z ≤ z) = 0.755 is 0.69.
The lower bound is,
and the upper bound is
.
Answer:
i hope you can see it
Step-by-step explanation:
Use Symbolab I use it all the time it gives correct answer and it gives good explanations (this isn't a sponsor I've just been using it since the 6th grade lol)