What is 10+30(39)x when x equals 3

Please help me solve all these problems 1-9 thanks in advanced

Elodia [21]

1) 50:1
2) 50:1
3) 49:1
4) 16:1
5) 30:1

8 0

3 years ago

Can anyone help me please

snow_lady [41]

Answer:

a. 41.5

b, 26.5

Step-by-step explanation:

a.

2x + 6 = 180

2x = 172

x = 86

b.

2x+ 90 +26+11 = 180

2x+127 = 180

2x = 53

x = 26.5

5 0

3 years ago

Read 2 more answers

Keegan deposits into her savings account at the beginning of the year. The account earns 3% simple interest each year. She has $

fiasKO [112]

Answer:

She deposits $350 into the account at the beginning of the year.

Step-by-step explanation:

Let us assume that Keegan deposits $P into her savings account at the beginning of the year.

The account earns 3% simple interest each year and she has $360.50 in her account at the end of the year.

If Keegan did not make any additional deposits or withdrawals during the year.

So, we can write that $P[1+\frac{3}{100} ]= 360.50$

⇒ P = $350 (Answer)

7 0

3 years ago

Tracy ran a lap around the school track in 74.2 seconds.Malcom ran a lap in 65.92 seconds. Estimate the difference in the times

Nataly_w [17]

Answer:

The difference would be 8.3 second

Step-by-step explanation:

5 0

4 years ago

An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem

Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

Case 1: both balls are white.

At the beginning we have $b+w$ balls. We want to pick a white one, so we have a probability of $\frac{w}{b+w}$ of picking a white one.

If this happens, we're left with $w-1$ white balls and still $b$ black balls, for a total of $b+w-1$ balls. So, now, the probability of picking a white ball is

$\dfrac{w-1}{b+w-1}$

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

$\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}$

Case 2: both balls are black

The exact same logic leads to a probability of

$\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}$

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

$\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of $\frac{w}{b+w}$ of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability $\frac{b}{b+w}$ of picking a black ball with both picks.

This leads to an overall probability of

$\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}$

Of picking two balls of the same colour.

We want to prove that

$\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Expading all squares and products, this translates to

$\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}$

As you can see, this inequality comes in the form

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k}$

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y$

And this is our case, because in our case we have

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

4 0

3 years ago