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pickupchik [31]

3 years ago

14

How do u write a ratio

1 answer:

Vsevolod [243]3 years ago

4 0

You can write a ratio like this:
Compare 5 oz of water to 19 oz of paint
Three types of ratios:
5/19, 5 to 19, 5:19

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PLEASE I'm DESPERATE!. Find a, b, c, and d such that the cubic . f(x) = ax3 + bx2 + cx + d. satisfies the given conditions.. Rel

dimulka [17.4K]

We are given with the equation <span>f(x) = ax3 + bx2 + cx + d

Substituting, (3,11)
</span><span>11= 27a + 9b + 3c + d
</span><span>@(5, 9)
</span><span>9 = 125 a + 25 b + 5c + d
@</span><span>(4, 10)
</span><span>10 = 64 a + 16 b + 4c + d

@inflection point, second derivative is equal to zero
</span><span>f'(x) = 3ax2 + 2bx + c
</span>f''(x) = 6ax + 2b = 0

when x is 4, 24 a + 2b = 0 or 12a + b = 0.

There are 4 equations, 4 unknowns: answer is
<span>0.5 x^3 - 6x^2 + 22.5 - 24 = 0</span>

4 0

3 years ago

Read 2 more answers

A scale drawing for a backyard is shown below.

scZoUnD [109]

The area of real patio is 50m².

Step-by-step explanation:

Given,

Length of patio = 3 cm

Width of patio = 6 cm

According to scale =

3 cm = 5m

1 cm = $\frac{5}{3}\ m$

6 cm = $\frac{5}{3}*6 = 10\ m$

Real length of patio = 5 m

Real width of patio = 10 m

Area = Length*width

$Area=5*10\\Area=50\ m^2$

The area of real patio is 50m².

Keywords: area, multiplication

Learn more about areas at:

#LearnwithBrainly

3 0

3 years ago

Jose has scored 562 points on his math tests so far this semester he must score atleast 650 points write and solve an inequality

solniwko [45]

562 + X >= 650
subtract 562 from both sides
X >= 88

5 0

3 years ago

Read 2 more answers

Suppose you are given either a fair dice or an unfair dice (6-sided). You have no basis for considering either dice more likely

hoa [83]

Answer: Our required probability is 0.83.

Step-by-step explanation:

Since we have given that

Number of dices = 2

Number of fair dice = 1

Probability of getting a fair dice P(E₁) = $\dfrac{1}{2}$

Number of unfair dice = 1

Probability of getting a unfair dice P(E₂) = $\dfrac{1}{2}$

Probability of getting a 3 for the fair dice P(A|E₁)= $\dfrac{1}{6}$

Probability of getting a 3 for the unfair dice P(A|E₂) = $\dfrac{1}{3}$

So, we need to find the probability that the die he rolled is fair given that the outcome is 3.

So, we will use "Bayes theorem":

$P(E_1|A)=\dfrac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\\\\(E_1|A)=\dfrac{0.5\times 0.16}{0.5\times 0.16+0.5\times 0.34}\\\\P(E_1|A)=0.83$

Hence, our required probability is 0.83.

8 0

3 years ago

In algebra, why do you think that you are given many problems to solve that aren't really "real life" problems? For example, "Tw

Liula [17]

Answer:

Ummm because it makes you smarter???? I feel like this is a question to ask your teacher haha

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers