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2 years ago

I need help pls help

Mathematics

1 answer:

Andreyy892 years ago

7 0

Answer:

[?] = 4

(the complete term is $4\frac{1}{2}\ \text{cups}$ )

Step-by-step explanation:

$8\ \text{cups} - 1\frac{1}{3}\ \text{cups} - 2\frac{1}{6}\ \text{cups} = (8-1-2)\ \text{cups} + \frac{-\frac{6}{3} - 1}{6}\ \text{cups} = 5\ \text{cups} + \frac{-3}{6}\ \text{cups} = 5\ \text{cups} + \frac{-1}{2}\ \text{cups} = 4\ \text{cups} + \frac{2 - 1}{2}\ \text{cups}$

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Maria is comparing the prices of two window cleaning companies. Company A charges $6 per window and an additional $12 as service

bija089 [108]

Part A
y=6x+12
y=5x+15
x is the amount of windows, y is the total amount of money
Part B
y=6(8)+12 =60
y=5(8)+15 =52
Company B
Part C
y=6(6)+12 =48
y=5(6)+12 =42
$6 is saved

4 0

3 years ago

A chemical plant has an emergency alarm system. When an emergency situation exists, the alarm sounds with probability 0.95. When

Lapatulllka [165]

Answer:

6.56% probability that a real emergency situation exists.

Step-by-step explanation:

We have these following probabilities:

A 0.4% probability that a real emergency situation exists.

A 99.6% probability that a real emergency situation does not exist.

If an emergency situation exists, a 95% probability that the alarm sounds.

If an emergency situation does not exist, a 2% probability that the alarm sounds.

The problem can be formulated as the following question:

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:

What is the probability of a real emergency situation existing, given that the alarm has sounded.

P(B) is the probability of there being a real emergency situation. So $P(B) = 0.004$ .

P(A/B) is the probability of the alarm sounding when there is a real emergency situation. So P(A/B) = 0.95.

P(A) is the probability of the alarm sounding. This is 95% of 0.4%(real emergency situation) and 2% of 99.6%(no real emergency situation). So

P(A) = 0.95*0.04 + 0.02*0.996 = 0.05792

Given that the alarm has just sounded, what is the probability that a real emergency situation exists?

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.004*0.95}{0.05792} = 0.0656$

6.56% probability that a real emergency situation exists.

6 0

3 years ago

How many solutions does this equation have

larisa86 [58]

Answer:

Two solutions

Step-by-step explanation:

This is a quadratic equation in the form y = ax² + bx + c.

For quadratic equations, you can find solutions using the quadratic formula:

$x = \frac{-b±\sqrt{b^{2}-4ac}}{2a}$ .

To find the number of solutions, you only need what's inside the square root. We call it the "discriminant" because lets us know the number of solutions without solving.

$b^{2}-4ac$

If b²- 4ac > 0, two solutions. (greater than)

If b²- 4ac < 0, no solutions. (less than)

If b²- 4ac = 0, one solution. (equal to)

y = ax² + bx + c

y = -3x² + x + 12

a = -3 b = 1 c = 12

Substitute into the discriminant

b²- 4ac

= 1² - 4(-3)(12)

= 1 - (-144)

= 145 > 0

b²- 4ac > 0 Discriminant greater than 0

Therefore, there are two solutions.

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3 years ago

Prove your very brainly ;)

scZoUnD [109]

Answer:

can you show what it says better its blurry

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4 years ago

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Answer this question please and thank you

Andrews [41]

7^3
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3 years ago

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