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2 years ago

How do you solve (4c-10)◙

Mathematics

1 answer:

Ratling [72]2 years ago

5 0

What is the black box? and u solve that by forgetting the c is there so 4-10= -6 or -6c

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HELP ASAP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

miv72 [106K]

Answer:

Step-by-step explanation:

The volume (V) of water in the tank is calculated as

V = lbh (l is length, b is breadth and h is height )

Here l = 80, b = 40 and h = 50 , then

V = 80 × 40 × 50 = 160000 cm³

V ( in litres) = 160000 ÷ 1000 = 160 litres , then

number of times = 160 ÷ 10 = 16

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2 years ago

Steve ran a distance of 150 meters in 1 1/2 minutes. What is his speed in meters per<br> hour?

WITCHER [35]

Answer:

600/mph

Step-by-step explanation:

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2 years ago

Can somebody please explain this to me? it makes no sense

RSB [31]

Answer: 33

Step-by-step explanation: So, each box is 11 pages, how do you we know that? Because it says there's 22 pages of illustrations...22 divided by 2 = 11.

Now that we know that, we know there's 33 pages of text because there's one more box and each box is 11 pages. Therefore 3 boxes = 33 pages. That means the ratio is 3:2. There's 3 boxes of texts and 2 boxes of illustrations, therefore 3:2 Hope this helps!

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3 years ago

Which expressions are equivalent to 5x - 15? Select three options.

Zigmanuir [339]

Answer:

-5(x-3)

-5(3-x)

-3(-5+2x)+x

Step-by-step explanation:

8 0

3 years ago

A new, simple test has been developed to detect a particular type of cancer. The test must be evaluated before it is used. A med

g100num [7]

Answer:

0.3333 = 33.33% probability of a randomly chosen person not having cancer given that the test indicates cancer

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Test indicates cancer.

Event B: Person does not have cancer.

Probability of a test indicating cancer.

98% of 2%(those who have).

1% of 100 - 2 = 98%(those who do not have). So

$P(A) = 0.98*0.02 + 0.01*0.98 = 0.0294$

Probability of a test indicating cancer and person not having.

1% of 98%. So

$P(A \cap B) = 0.01*0.98 = 0.0098$

What is the probability of a randomly chosen person not having cancer given that the test indicates cancer?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0098}{0.0294} = 0.3333$

0.3333 = 33.33% probability of a randomly chosen person not having cancer given that the test indicates cancer

3 0

3 years ago