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3 years ago

A researcher read a 12-page article in 30 minutes. On

1 answer:

3 0

Answer: 2.5 minutes

Explanation: Divide the time by the amount of pages to find the amount of time spent of each page. (30/12=2.5)

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Evaluate 2.43+51.62+124.77

Damm [24]

2.43 + 1.62 + 124.77 = 178.82 Because all you have to do is add up all the numbers (Don't take the decimals out, leave them in) And then you get: 178.82.

Hope I helped!

- Amber

3 0

2 years ago

20/35 as a fraction in simplest form

zheka24 [161]

$\frac{20}{35}=\frac{4 \times 5}{7 \times 5}=\boxed{\frac{4}{7}}$

7 0

3 years ago

Read 2 more answers

Solve |x| + 7 < 4. ........................

Fynjy0 [20]

Hello :
|x| + 7 < 4.
|x| < 4.-7
|x| < -3
no reals solutions because for all reals x : |x| ≥ 0

8 0

3 years ago

If log a = x and log b = y, then log (ab2) equals

mylen [45]

Answer:

The answer for log(ab²) is x + 2y.

Step-by-step explanation:

You have to apply Logarithm Law,

$log(a) + log(b) \: ⇒ \: log(a \times b)$

$log( {a}^{n} ) \: ⇒ \: n log(a)$

In this question, you have to seperate it out :

$log(a {b}^{2} ) = log(a) + log( {b}^{2} )$

$log( {b}^{2} ) = 2 log(b)$

$let \: log(a) = x$

$let log(b) = y$

$log(a {b}^{2} ) = log(a) + 2 log(b)$

$log(a {b}^{2} ) = x + 2y$

3 0

3 years ago

A used car dealer has 30 cars and 10 of them are lemons (i.e.~ mechanically faulty used cars), and you don't know which is which

Triss [41]

Answer: Hence, the probability that he will get at least one lemon is 0.70.

Step-by-step explanation:

Since we have given that

Number of cars = 30

Number of lemon cars = 10

Number of other than lemon cars = 30-10 = 20

According to question, he bought 3 cars,

we need to find the probability that you will get at least one lemon.

So, P(X≤1)=1-P(X=0)=1-P(no lemon)

Here, P(no lemon ) is given by

$\dfrac{20}{30}\times \dfrac{20}{30}\times \dfrac{20}{30}=(\dfrac{20}{30})^3$

so, it becomes,

$P(X\geq 1)=1-(\dfrac{20}{30})^3=1-(0.67)^3=0.70$

Hence, the probability that he will get at least one lemon is 0.70.

5 0

3 years ago