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4 years ago

A chili recipe calls for 6 lb of ground beef for 25 servings how many pounds of ground beef are needed for 30 servings

Mathematics

1 answer:

Vika [28.1K]4 years ago

8 0

4lbs of ground beef should be the answer

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Which is the best estimate for the average rate of change for the quadratic function graph on the interval 0 ≤ x ≤ 4?

den301095 [7]

ANSWER

A) -1

EXPLANATION

The average rate of change of the given quadratic function on the interval 0 ≤ x ≤4 is the slope of the secant line connecting the points (0,f(0)) and (4,f(4)).

That is the average rate of change is:

$= \frac{f(4) - f(0)}{4 - 0}$

From the graph, f(0) is 0 and f(4) is -4.

We plug in these values to obtain;

$= \frac{ - 4 - 0}{4 - 0}$

This simplifies to;

$= \frac{ - 4}{4}$

$= - 1$

Hence the average rate of change for the given quadratic function whose graph is shown on 0≤x≤4 is -1

5 0

3 years ago

Read 2 more answers

Which expression represent the distance between the point graphed below

agasfer [191]

The answer is A because when you plug in the points into the distance formula, you will get square root of 17 squared plus 17 squared. Hope this helps!

5 0

3 years ago

If y has moment-generating function m(t) = e 6(e t −1) , what is p(|y − µ| ≤ 2σ)?

Lostsunrise [7]

Since $\mathrm M_Y(t)=e^{6(e^t-1}$ , we know that $Y$ follows a Poisson distribution with parameter $\lambda=6$ .

Now assuming $\mu,\sigma$ denote the mean and standard deviation of $Y$ , respectively, then we know right away that $\mu=6$ and $\sigma=\sqrt6$ .

So,

$\mathbb P(|Y-\mu|\le2\sigma)=\mathbb P(6-2\sqrt6\le Y\le6+2\sqrt6)=\dfrac{66366}{175e^6}\approx0.940028$

6 0

4 years ago

Help as soon as possible<br> The greatest common factor of y^4. y^6 and y^5 is

olchik [2.2K]

Answer:

Answer:GCF: y^4

Step-by-step explanation:

Factors of:

y^4= (y) (y) (y) (y) = y^4 (1)

y^5 = (y) (y) (y) (y) (y) = y^4 (y)

y^6 = (y) (y) (y) (y) (y) (y) = y^4(y^2)

6 0

2 years ago

Question 2 (1 point)

Papessa [141]

Answer:

55.734 mph

Step-by-step explanation:

Let's call the wind speed x and the time of flight t.

Then, we can write the following equation, knowing that the distance is equal speed times time of flight:

910 = (350+x) * t

660 = (350-x) * t

If we isolate t in both equations and compare its value, we have that:

910 / (350+x) = 660 / (350-x)

(350+x)/(350-x) = 910/660

350 + x = 1.3788 * (350-x)

350 + x = 482.58 - 1.3788x

2.3788x = 132.58

x = 55.734 mph

4 0

4 years ago