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3 years ago

10

How to solve for x on a quadratic function

1 answer:

Sliva [168]3 years ago

7 0

To find the x<span>-intercepts of any equation, substitute 0 in for y and </span>solve for x. So, we have 0 = 3x2<span>+ </span>x<span> + 1. Now, use the </span>quadratic<span> equation to </span>solve for x<span>, wich a = 3, b = 1, and c = 1: So, now we can find the value of the </span>x<span>-intercepts and not have to estimate!</span>

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Can someone help me with number 1 and 2? I just need to know what to do please!!

max2010maxim [7]

1) is 1 2) 2 is 8.60 4 is 10.00 5 is 24.00

6 0

4 years ago

Whats the area of the circle above

Anna35 [415]

Answer:

<h2>The answer is option A</h2>

Step-by-step explanation:

Area of a circle = πr²

where

r is the radius of the circle

From the question

r = 15.2 cm

The area of the circle is

A = (15.2)²π

= 231.04π

= 725.833566685386...

We have the final answer as

<h3>725.8 cm²</h3>

Hope this helps you

7 0

3 years ago

The gross weekly sales at a certain restaurant are a normal random variable with mean $2200 and standard deviation $230.What is

Lesechka [4]

Answer:

a) The total gross sales over the next 2 weeks exceeds $5000 is 0.0321.

b) The weekly sales exceed $2000 in at least 2 of the next 3 weeks is 0.9033.

Step-by-step explanation:

Given : The gross weekly sales at a certain restaurant are a normal random variable with mean $2200 and standard deviation $230.

To find : What is the probability that

(a) the total gross sales over the next 2 weeks exceeds $5000;

(b) weekly sales exceed $2000 in at least 2 of the next 3 weeks? What independence assumptions have you made?

Solution :

Let $X_1$ and $X_2$ denote the sales during week 1 and 2 respectively.

a) Let $X=X_1+X_2$

Assuming that $X_1$ and $X_2$ follows same distribution with same mean and deviation.

$E(X)=E(X_1+X_2)=E(X_1)+E(X_2)$

$E(X)=2\mu = 2(220)=4400$

$\sigma_X=\sqrt{var(X_1+X_2)}$

$\sigma_X=\sqrt{2\sigma^2}$

$\sigma_X=\sqrt{2}\sigma$

$\sigma_X=230\sqrt{2}$

So, $X\sim N(4400,230\sqrt{2})$

$P(X>5000)=1-P(X\leq5000)$

$P(X>5000)=1-P(Z\leq\frac{5000-4400}{230\sqrt{2}})$

$P(X>5000)=1-P(Z\leq1.844)$

$P(X>5000)=1-0.967$

$P(X>5000)=0.0321$

The total gross sales over the next 2 weeks exceeds $5000 is 0.0321.

b) The probability that sales exceed teh 2000 and amount in at least 2 and 3 next week.

We use binomial distribution with n=3.

$P(X>2000)=1-P(X\leq2000)$

$P(X>2000)=1-P(Z\leq\frac{2000-2200}{230})$

$P(X>2000)=1-P(Z\leq-0.87)$

$P(X>2000)=1-0.1922$

$P(X>2000)=0.808$

Let Y be the number of weeks in which sales exceed 2000.

Now, $P(Y\geq 2)$

So, $P(Y\geq 2)=P(Y=2)+P(Y=3)$

$P(Y\geq 2)=^3C_2(0.8077)^2\cdot (1-0.8077)+^3C_3(0.8077)^3$

$P(Y\geq 2)=0.37635+0.52692$

$P(Y\geq 2)=0.90327$

The weekly sales exceed $2000 in at least 2 of the next 3 weeks is 0.9033.

3 0

3 years ago

A bicycle is originally priced at $80 . The store owner gives a discount and the bicycle is now priced at $60 . Enter the percen

Varvara68 [4.7K]

If the bike cost $80 and got it with a discount to $60 the answer is 20%

8 0

3 years ago

Which statement(s) are true?

Anika [276]

Answer:

no event can be impossible

4 0

2 years ago

Read 2 more answers