Divide. Check your answer.

A waether forecast predicts a 30% chance of rain for each of the next 3 days. Describe a way to stimulate the chance that it wil

DerKrebs [107]

Using a binomial distribution considering there's a 30% chance it will rain on any of the three days:

The probability of it raining on 0 days is (1)(0.7)(0.7)(0.7) = 34.3%. 
The probability of it raining on 1 day is (3)(0.3)(0.7)(0.7) = 44.1%. 
The probability of it raining on 2 days is (3)(0.3)(0.3)(0.7) = 18.9%. 
The probability of it raining on 3 days is (1)(0.3)(0.3)(0.3) = 2.7%. 

There's a 65.7% chance that it will rain at least once over the three-day period.

5 0

3 years ago

Please Help Me! I need to pass.

Andru [333]

42 is the answer

Explanation:
50% of 80 is half of 80 which is 40

Then you would add 40 with 5% of 40

What you would do then is 5% times 40 then you would change the 5% into a decimal which is 0.05 and multiply it by 40 , which would get u 2

Then you would add the price with tax
40+2 = 42

6 0

3 years ago

You saved $300 to spend over the summer. You decide to budget $50 to spend each week.

Oduvanchick [21]

Answer:

Part c)

The Slope of the line is: m=-50 and represents the amount of money spent per week.

Part d)

The y-intercept is: c=300 and represents the maximum money we have that can be spend over the weeks (i.e. our maximum budget alowed).

Step-by-step explanation:

To solve this question we shall look at linear equations of the simplest form reading:

$y = mx+c$ Eqn(1).

where:

$y$ : is our dependent variable that changes as a function of x

$x$ : is our independent variable that 'controls' our equation of y

$m$ : is the slope of the line

$c$ : is our y-intercept assuming an $x$ ⇔ $y$ relationship graph.

This means that as $x$ changes so does $y$ as a result.

Given Information:

Here we know that $300 is our Total budget and thus our maximum value (of money) we can spend, so with respect to Eqn (1) here:

$c=300$

The budget of $50 here denotes the slope of the line, thus how much money is spend per week, so with respect to Eqn (1) here:

$m=50$

So finally we have the following linear equation of:

$y= - 50x + 300$ Eqn(2).

Notice here our negative sign on the slope of the line. This is simply because as the weeks pass by, we spend money therefore our original total of $300 will be decreasing by $50 per week.

So with respect to Eqn(2), and different weeks thus various $x$ values we have:

Week 1: $x=1$ we have $y= -50 *1 + 300 = -50 +300 = 250$ dollars.

Week 2: $x=2$ we have $y= -50 *2 + 300 = -100 +300 = 200$ dollars.

Thus having understood the above we can comment on the questions asked as follow:

Part c)

The Slope of the line is: $m=-50$ and represents the amount of money spent per week.

Part d)

The y-intercept is: $c=300$ and represents the maximum money we have that can be spend over the weeks (i.e. our maximum budget alowed).

4 0

3 years ago

If m 1 = 90°, what is m?

Setler79 [48]

Perpendicular

Mark as brainliest

7 0

4 years ago

Read 2 more answers

9.21 <img src="https://tex.z-dn.net/?f=9.21%20%5Cdiv%200.4%20%3D%20" id="TexFormula1" title="9.21 \div 0.4 = " alt="9.21 \di

Whitepunk [10]

Let us do it by converting the 0.4 to a fraction, it has one decimal, so we'll use one zero at the denominator.

and 9.21 to a fraction as well, two decimals, thus two zeros at the denominator then,

$\bf 9.21\div 0.4\\\\
-------------------------------\\\\
0.\underline{4}\implies \cfrac{04}{1\underline{0}}\implies \cfrac{4}{10}\implies \cfrac{2}{5}\qquad \qquad \qquad \qquad 9.\underline{21}\implies \cfrac{921}{1\underline{00}}\\\\
-------------------------------\\\\
9.21\div 0.4\implies \cfrac{921}{100}\div \cfrac{2}{5}\implies \cfrac{921}{100}\cdot \cfrac{5}{2}\implies \cfrac{921}{20}\cdot \cfrac{1}{2}\implies \cfrac{921}{40}
\\\\\\
23\frac{1}{40}\implies 23.025$

3 0

4 years ago