The graph of the function B is shown below. If B(x) = -1, then what is x?

How do i solve this because I need some help. I'm studying for my test.

mote1985 [20]

At Gino's you pay $16 plus $8 per pizza.
At Venetian's you pay $24 plus $6 per pizza.
Let the number of pizzas be x.

At Gino's you pay 16 + 8x
At Venetian's you pay 24 + 6x

Set the two costs equal and solve for x to find out the number of pizzas for which both costs are the same.

16 + 8x = 24 + 6x
16 + 2x = 24
2x = 8
x = 4
Each place gives you a free pizza.
4 pizzas plus the free pizza equals 5 pizzas.

If you need 5 pizzas (including the free one), both parlors cost the same.

If you need fewer than 5 pizzas, use Gino's.
If you need more than 5 pizzas, use Venetian's.

7 0

3 years ago

Evaluate the limit assuming that limx→−5f(x)=17limx→−5f(x)=17 and limx→−5g(x)=22limx→−5g(x)=22. (use symbolic notation and fract

Gemiola [76]

In this question it is given that

$\lim_{x->-5}f(x)=17, \lim_{x->-5}g(x)=22$

And we have to find the value of the given limit

$\lim_{x->-5}(23f(x)+3g(x))$

Using properties of limit, first we separate the two functions, that is

$23\lim_{x->-5}f(x)+3\lim_{x->-5}g(x)$

Substituting the values of the given limit,

$23(17)+3(22)=457$

3 0

3 years ago

Question 6 The mineral content of a particular brand of supplement pills is normally distributed with mean 490 mg and variance o

AysviL [449]

Answer:

0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean 490 mg and variance of 400.

This means that $\mu = 490, \sigma = \sqrt{400} = 20$