Find the indicated side of the triangle. a 7. 30° D b = [ ? IV

Question 1 of 10 What is the x-intercept of the function graphed below?

aleksley [76]

The "x-interept" is the point where the graph crosses the x-axis.

From here, it looks like this graph crosses the x-axis where x=2 .

So the point is (2, 0) .

That's choice-A .

8 0

3 years ago

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On Tuesday, a local hamburger shop sold a combined total of 576 hamburgers and cheeseburgers. The number of cheeseburgers sold w

cestrela7 [59]

Answer:

check the picture

Step-by-step explanation:

I hope this helps

4 0

3 years ago

Given f(x)= √7x and g(x) = 1/x-8 which value is in the domain of f^o g

bekas [8.4K]

Answer: x>8

Step-by-step explanation: when we plug g(x) into f(x) we get sqrt7(1/x-8) and when you graph (as shown below) this you see that the graph has a vertical asymptote of 8 therefore the domain is x>8

8 0

3 years ago

If we sample from a small finite population without replacement, the binomial distribution should not be used because the event

seropon [69]

Answer:

5/4324 = 0.001156337

Step-by-step explanation:

To better understand the hyper-geometric distribution consider the following example:

There are 100 senators in the US Congress, and suppose 60 of them are republicans so 100 - 60 = 40 are democrats).

We extract a random sample of 30 senators and we want to answer this question:

What is the probability that 10 senators in the sample are republicans (and of course, 30 - 10 = 20 democrats)?

The answer using the h-g distribution is:

$\large \frac{\binom{60}{10}\binom{100-60}{30-10}}{\binom{100}{30}}=\frac{\binom{60}{10}\binom{40}{20}}{\binom{100}{30}}$

Now, imagine there are 56 senators (56 lottery numbers), 6 are republicans (6 winning numbers and 50 losers), we extract a sample of 6 senators (the bettor selects 6 numbers). What is the probability that 4 senators are republicans? (What is the probability that 4 numbers are winners?).

As we see, the situation is exactly the same, but changing the numbers. So the answer would be

$\large \frac{\binom{6}{4}\binom{56-6}{6-4}}{\binom{56}{6}}=\frac{\binom{6}{4}\binom{50}{2}}{\binom{56}{6}}$

Now compute each combination separately:

$\large \binom{6}{4}=\frac{6!}{4!2!}=15\\\\\binom{50}{2}=\frac{50!}{2!48!}=1225\\\\\binom{50}{6}=\frac{50!}{6!44!}=15890700$

and now replace the values:

$\large \frac{\binom{6}{4}\binom{50}{2}}{\binom{56}{6}}=\frac{15*1225}{15890700}=\frac{18375}{15890700}=\frac{5}{4324}$

and that is it.

If the decimal expression is preferred then divide the fractions to get 0.001156337

6 0

3 years ago

All sacks of sugar have the same weight. All sacks of flour also have the same weight, but not necessarily the same as the weigh

BaLLatris [955]

Answer:

The largest possible weight of flour is 11.25 pounds.

Step-by-step explanation:

To start with, we will assume that the weight of 1 sack of sugar = x pounds

We will also assume that the weight of 1 sack of flour = y pounds

So, the weight of 2 sacks of sugar = 2 * (x) = 2x

Same thing goes for the weight of 3 sacks of flour = 3 * (y) = 3y

Supposing that the weight of (2 sacks of sugar + 3 sacks of flour) ≤ 40 pounds

= 2x + 3y ≤ 40............ we'll call that equation 1.

Also, suppose that the weight of ( 1 sack of flour) ≤ 2 sacks of sugar + 5 pounds

= y ≤ 2x + 5........................ we'll call that equation 2

Therefore, we'll solve for the values of x and y in the two equations and we will get:

2x + 3y ≤ 40

y ≤ 2x + 5

Now, substitute the value of y into equation 1

2x + 3y ≤ 40 ⇒ 2x + 3(2x +5) =40

⇒ 2x + 6x + 15= 40

⇒ 8x + 15 = 40

⇒ 8x = 25

⇒ x = 25/8

⇒ x = 3.12

x cannot be more than 3.12 pounds, so we solve for y

Putting the value of x into equation 2, we'll get

⇒ 2y + 5 = 2(3.12) + 5

⇒ y = 11.25 pounds.

So, n cannot be more than 11.25 pounds

8 0

3 years ago