All categories

2 years ago

A. If the average ticket price is $15

Mathematics

1 answer:

Westkost [7]2 years ago

5 0

Answer:

He had been in a hospital with friends since his birth

You might be interested in

I need help with my math

kherson [118]

Answer:

Ok what is the problem

Step-by-step explanation:

8 0

2 years ago

Which of the following is an equation for the plane through the point (1, 1, −1) and parallel

Bas_tet [7]

The given plane, $x-y+z=3$ , has normal vector $\vec n = \langle1,-1,1\rangle$ . Any plane parallel to this one has the same normal vector.

Let $(x,y,z)$ be any point in the plane we want. The plane contains the point (1, 1, -1), so an arbitrary vector in this plane is

$\langle x,y,z\rangle - \langle 1,1,-1\rangle = \langle x-1, y-1, z+1 \rangle$

and this is perpendicular to $\vec n$ .

So the equation of the plane is

$\langle x-1, y-1, z+1 \rangle \cdot \vec n = 0 \implies (x-1) - (y-1) + (z+1) = 0 \implies \boxed{x - y + z = -1}$

or equivalently,

$\boxed{-x + y - z = 1}$

6 0

1 year ago

Use foil (a + 3)(a - 2)

blsea [12.9K]

Answer:

$\left(a+3\right)\left(a-2\right)=a^2+a-6=$

Step-by-step explanation:

Given expression is $\left(a+3\right)\left(a-2\right)$ .

Now we need to multiply this using FOIL.

F = First $=\left(a\right)\left(a\right)= a^2$

O = Outside $=\left(a\right)\left(-2\right)= -2a$

I = Inside $=\left(3\right)\left(a\right)= 3a$

L = Last $=\left(3\right)\left(-2\right)= -6$

Hence we get :

$\left(a+3\right)\left(a-2\right)=a^2-2a+3a-6=a^2+a-6=$

7 0

3 years ago

Read 2 more answers

Help here too please

lyudmila [28]

What’s the question

7 0

3 years ago

An electronics company just finished designing a new tablet computer and is interested in estimating its battery-life. A random

Alika [10]

Answer:

$6 \pm 2.861(0.3354)$

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 20 - 1 = 19

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 2.861.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.861\frac{1.5}{\sqrt{20}} = 2.861(0.3354)$

In which s is the standard deviation of the sample and n is the size of the sample.

The format of the confidence interval is:

$S_{M} \pm M$

In which $S_{M}$ is the sample mean

$6 \pm 2.861(0.3354)$

7 0

2 years ago