All categories

3 years ago

Please help me with this I need it bad

Mathematics

1 answer:

lana66690 [7]3 years ago

3 0

Answer:

Step-by-step explanation:

The average rate of change is the gradient _

Gradient = Rise / Run = (y2 - y1) / (x2 - x1)

From the table ;

x2 = 4 ; x1 = - 1 ; y2 = 5 ; y1 = 0

Gradient = (y2 - y1) / (x2 - x1) = (5-0) / (4 - - 1) = 5/5 = 1

You might be interested in

HEY THERE! PLEASE ANSWER THE MATH QUESTION IN THE PHOTO! THE CORRECT ANSWER WILL BE MARKED AS BRAINLIEST ,GET POINTS AND I WILL

Bad White [126]

Answer:

a) 1/10

b) 2/10

c) 8/10

8 0

3 years ago

Read 2 more answers

To reduce laboratory costs, water samples from two public swimming pools are combined for one test for the presence of bacteria

givi [52]

Answer:

The probability of a positive test result is 0.017919

Option C is correct.

The probability is quite low, indicating that further testing of the individual samples will be a rarely necessary event.

Step-by-step explanation:

Probability of finding bacteria in one public swimming pool = 0.009

We now require the probability of finding bacteria in the combined test of two swimming pools. This probability is a sum of probabilities.

Let the two public swimming pools be A and B respectively.

- It is possible for public swimming pool A to have bacteria and public swimming pool B not to have bacteria. We would obtain a positive result from testing a mixed sample of both public swimming pools.

- It is also possible for public swimming pool A to not have bacteria and public swimming pool B to have bacteria. We would also obtain a positive result from testing a mixed sample of both public swimming pools.

- And lastly, it is possible that both swimming pools both have bacteria in them. We will definitely get a positive result from this too.

So, if P(A) is the probability of the event of bacteria existing in public swimming pool A

And P(B) is the probability of the event of bacteria existing in public swimming pool B

P(A') and P(B') represent the probabilities of bacteria being absent in public swimming pool A and public swimming pool B respectively.

P(A) = P(B) = 0.009

P(A') = P(B') = 1 - 0.009 = 0.991

Since the probabilities for each public swimming pool is independent of the other.

P(A or B) = P(A n B') + P(A' n B) + P(A n B)

= P(A)×P(B') + P(A')×P(B) + P(A)×P(B)

= (0.009×0.991) + (0.991×0.009) + (0.009×0.009)

= 0.008919 + 0.008919 + 0.000081

= 0.017919

Evidently, a probability of 0.017919 (1.7919%) indicates an event with a very low likelihood. A positive result is expected only 1.7919% of the time.

Hence, we can conclude that the probability is quite low, indicating that further testing of the individual samples will be a rarely necessary event.

Hope this Helps!!!

4 0

4 years ago

Read 2 more answers

How to find the area of the compound figure

Oksana_A [137]

The total area is 84m.

Work is in the image, feel free to ask if you don’t understand it.

Hope this helped :)

3 0

3 years ago

Which transformations can be used to carry ABCD onto itself? The point of rotation is (3, 2). Check all that apply. A. Reflectio

Blizzard [7]

The answer would be B and D

4 0

4 years ago

In evaluating a double integral over a region D, a sum of iterated integrals was obtained as follows:

BabaBlast [244]

Answer

$a=0$ , $b=2$

$g_1(x)=\frac{5x}{2}$ , $g_2(x)=7-x$

Step-by-step explanation:

Given that

$\int \int Df(x,y)dA=\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy+\int_5^7\int_0^{7-y} f(x,y)dxdy\; \cdots (i)$

For the term $\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy$ .

Limits for $x$ is from $x=0$ to $x=\frac {2y}{5}$ and for $y$ is from $y=0$ to $y=5$ and the region $D$ , for this double integration is the shaded region as shown in graph 1.

Now, reverse the order of integration, first integrate with respect to $y$ then with respect to $x$ . So, the limits of $y$ become from $y=\frac{5x}{2}$ to $y=5$ and limits of $x$ become from $x=0$ to $x=2$ as shown in graph 2.

So, on reversing the order of integration, this double integration can be written as

$\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy=\int_0 ^2\int _ {\frac {5x}{2}}^5 f(x,y)dydx\; \cdots (ii)$

Similarly, for the other term $\int_5 ^7\int _0 ^ {7-y} f(x,y)dxdy$ .

Limits for $x$ is from $x=0$ to $x=7-y$ and limits for $y$ is from $y=5$ to $y=7$ and the region $D$ , for this double integration is the shaded region as shown in graph 3.

Now, reverse the order of integration, first integrate with respect to $y$ then with respect to $x$ . So, the limits of $y$ become from $y=5$ to $y=7-x$ and limits of $x$ become from $x=0$ to $x=2$ as shown in graph 4.

So, on reversing the order of integration, this double integration can be written as

$\int_5 ^7\int _0 ^ {7-y} f(x,y)dxdy=\int_0 ^2\int _5 ^ {7-x} f(x,y)dydx\;\cdots (iii)$

Hence, from equations $(i)$ , $(ii)$ and $(iii)$ , on reversing the order of integration, the required expression is

$\int \int Df(x,y)dA=\int_0 ^2\int _ {\frac {5x}{2}}^5 f(x,y)dydx+\int_0 ^2\int _5 ^ {7-x} f(x,y)dydx$

$\Rightarrow \int \int Df(x,y)dA=\int_0 ^2\left(\int _ {\frac {5x}{2}}^5 f(x,y)+\int _5 ^ {7-x} f(x,y)\right)dydx$

$\Rightarrow \int \int Df(x,y)dA=\int_0 ^2\int _ {\frac {5x}{2}}^{7-x} f(x,y)dydx\; \cdots (iv)$

Now, compare the RHS of the equation $(iv)$ with

$\int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)dydx$

We have,

$a=0, b=2, g_1(x)=\frac{5x}{2}$ and $g_2(x)=7-x$ .

3 0

3 years ago