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3 years ago

1 1/4 + 3/6 pls help i need this i will give brianly i promise

Mathematics

1 answer:

Kazeer [188]3 years ago

4 0

Answer:

11/4+3/6 =13/4or 3 1/4

Step-by-step explanation:

(11*6) + (3*4)

4*6

78/24

78/6=13

24/6=4

13/4 or 3 1/4

13/4=3 1/4

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Suppose a particular type of cancer has a 0.9% incidence rate. Let D be the event that a person has this type of cancer, therefo

natita [175]

Answer:

There is a 12.13% probability that the person actually does have cancer.

Step-by-step explanation:

We have these following probabilities.

A 0.9% probability of a person having cancer

A 99.1% probability of a person not having cancer.

If a person has cancer, she has a 91% probability of being diagnosticated.

If a person does not have cancer, she has a 6% probability of being diagnosticated.

The question can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem we have the following question

What is the probability that the person has cancer, given that she was diagnosticated?

P(B) is the probability of the person having cancer, so $P(B) = 0.009$

P(A/B) is the probability that the person being diagnosticated, given that she has cancer. So $P(A/B) = 0.91$

P(A) is the probability of the person being diagnosticated. If she has cancer, there is a 91% probability that she was diagnosticard. There is also a 6% probability of a person without cancer being diagnosticated. So

$P(A) = 0.009*0.91 + 0.06*0.991 = 0.06765$

What is the probability that the person actually does have cancer?

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.91*0.009}{0.0675} = 0.1213$

There is a 12.13% probability that the person actually does have cancer.

3 0

3 years ago

The rectangular prism is to be sliced perpendicular to the shaded face and is to pass through point B, parallel to the front fac

kodGreya [7K]

The dimensions of the rectangular cross section will be 10 centimeters by 18 centimeters

Step-by-step explanation:

As ,we know

The rectangular cross section is parallel to the front face

Which clearly states that

The dimensions of the rectangular cross section is congruent with the dimensions of the front face

Lets assume that dimensions of the front face are 10 centimeters by 18 centimeters

Then ,The dimensions of the cross section will also be 10 centimeters by 18 centimeters

Hence we can say that the dimensions of the rectangular cross section will be 10 centimeters by 18 centimeters

4 0

3 years ago

NEED HELP i dont know how to do this and i need to turn in soon

bulgar [2K]

Answer:

f(-1) = 0 and f(4) = 0

Step-by-step explanation:

look at graph to see what point(s) is/are intersecting the y-axis, then use the y-value of the ordered pair

8 0

3 years ago

Find the area of the following parallelograms

Kay [80]

Answer:

Area of parallelogram: 1/2×h×sum of base lengths

Property of parallelogram: 2 parallel sides

Area = 1/2×4×(9+9)

= 36 cm^2

3 0

3 years ago

Read 2 more answers

Consider the function below, which has a relative minimum located at (-3 , -18) and a relative maximum located at (1/3, 14/17).

leonid [27]

First of all, when I do all the math on this, I get the coordinates for the max point to be (1/3, 14/27). But anyway, we need to find the derivative to see where those values fall in a table of intervals where the function is increasing or decreasing. The first derivative of the function is $f'(x)=-3x^2-8x+3$ . Set the derivative equal to 0 and factor to find the critical numbers. $0=-3x^2-8x+3$ , so x = -3 and x = 1/3. We set up a table of intervals using those critical numbers, test a value within each interval, and the resulting sign, positive or negative, tells us where the function is increasing or decreasing. From there we will look at our points to determine which fall into the "decreasing" category. Our intervals will be -∞<x<-3, -3<x<1/3, 1/3<x<∞. In the first interval test -4. f'(-4)=-13; therefore, the function is decreasing on this interval. In the second interval test 0. f'(0)=3; therefore, the function is increasing on this interval. In the third interval test 1. f'(1)=-8; therefore, the function is decreasing on this interval. In order to determine where our points in question fall, look to the x value. The ones that fall into the "decreasing" category are (2, -18), (1, -2), and (-4, -12). The point (-3, -18) is already a min value.

8 0

3 years ago