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3 years ago

Help -9x + 3.0 > 6.6 Group of answer choices x 0.4 x 2.4

Mathematics

1 answer:

Y_Kistochka [10]3 years ago

3 0

Answer:

x 0.4

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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Pls help due ASAP ....

Sati [7]

C because that’s where the lien is going

5 0

3 years ago

Which equation is not equivalent to the other two

Liula [17]

Answer:

y=-4-3x

Step-by-step explanation:

y = -3x - 4

y = -3x + (-4)

y = -4 + 3x

y = -4 - 3x

7 0

3 years ago

The tour bus had 60 passengers. The survey showed that 20% of the passengers enjoyed the tour. How many passengers enjoyed the t

aleksklad [387]

The answer is D. My dad taught me an easy way to solve those kind of problems.You need to do (60×20)÷100. And since 60 and 20 both have a zero and 100 has 2 zeros, u can cross out all the zeros and just do 6×2 and divide that by 1 and so 20% of 60 is 12. Hope this helped :)

5 0

3 years ago

Calculate the product of 8/15, 6/5, and 1/3.

Mariulka [41]

$\frac{8}{15} * \frac{6}{5} * \frac{1}{3}$

First, I would multiply the last two fractions because they're smaller and easier to work with:

$\frac{8}{15} * \frac{6}{5} * \frac{1}{3} = \frac{8}{15} * \frac{6}{15}$

Now that we've simplified it, we could multiply these terms and simplify. An easier method, however, would be to cancel out any common factors among the numerators and denominators before multiplying:

$\frac{8}{15} * \frac{6}{15} = \frac{8}{5} * \frac{2}{15}$

We can now multiply these terms:

$\frac{8}{5} * \frac{2}{15} = \frac{16}{75}$

The <span>product of 8/15, 6/5, and 1/3 is B, 16/75.</span>

5 0

3 years ago

Suppose that 8% of the general population has a disease and that the test for the diesease is accurate 70% of the time. What is

balu736 [363]

Answer:

P = 0.332

Step-by-step explanation:

The probability of having the disease is 0.08

The probability that the test predicts with accuracy is 0.7.

We need to find the probability that the test positive for the disease.

Several cases may occur.

Case 1.

You have the disease and the test predicts it accurately

$P_1 = 0.08(0.7) = 0.056$

Case 2

You do not have the disease and the test predicts that you have it

$P_2 = 0.92(0.3) = 0.276$

Then the probability that the test predicts that you have the disease is the union of both probabilities P1 and P2

$P = P_1 + P_2\\\\P = 0.056 + 0.276\\\\P = 0.332$

8 0

3 years ago