All categories

3 years ago

HELLLPPPPP!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

krek1111 [17]3 years ago

8 0

Answer:

1}6 times

Step-by-step explanation:

The highest number of president born in Virginia minus the highest number of <u>president</u> born in Texas

2}Yes.New York and Mast has the same number of president which is four

3}Yes,this can be done using piechart.

4}I am still working on this question

You might be interested in

Which decimal is equivalent to -3 1/8? *20 points*

kondaur [170]

Answer: -3.125

Step-by-step explanation: -1 divided by 8 = .125

5 0

3 years ago

An item is regularly priced at $24. Christine bought it on sale for 20% off the regular price. how much Christine paid.

AleksandrR [38]

$19.2, hope this helped

6 0

3 years ago

Read 2 more answers

Solve the system of equations using subtraction. <br> 2x - 4y = - 12 <br>2x + y = 13

alisha [4.7K]

First can you change
2x + y = -12
To have y as the subject

Looks like y = ???

7 0

3 years ago

In a recent year, a sample of grade 8 Washington State public school students taking a mathematics assessment test had a mean sc

Daniel [21]

Answer:

The number is $N =1147$ students

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 281$

The standard deviation is $\sigma = 34.4$

The sample size is n = 2000

percentage of the would you expect to have a score between 250 and 305 is mathematically represented as

$P(250 < X < 305 ) = P(\frac{ 250 - 281}{34.4 } < \frac{X - \mu }{\sigma } < \frac{ 305 - 281}{34.4 } )$

Generally

$\frac{X - \mu }{\sigma } = Z (Standardized \ value \ of \ X )$

$P(250 < X < 305 ) = P(-0.9012< Z$

$P(250 < X < 305 ) = P(z_2 < 0.698 ) - P(z_1 < -0.9012)$

From the z table the value of $P( z_2 < 0.698) = 0.75741$

and $P(z_1 < -0.9012) = 0.18374$

$P(250 < X < 305 ) = 0.75741 - 0.18374$

$P(250 < X < 305 ) = 0.57$

The percentage is $P(250 < X < 305 ) = 57\%$

The number of students that will get this score is

$N = 2000 * 0.57$

$N =1147$

6 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