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1 year ago

The probability of rain in April is 40%. If there are 30 days in April, how many days would you expect rain?

Mathematics

1 answer:

scoray [572]1 year ago

5 0

Answer:

12 days

Step-by-step explanation:

10% = 3 (30/10)

To find 40%, we can just times the above by 4.

3*4=12 therefore 12 days can expect rain.

Hope this helps!

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Think of your favorite spot to get food from. You are going there to get your first meal after quarantine and the meal cost $35.

sveta [45]

Answer:

You will be paying $41.30 in total.

Step-by-step explanation:

The cost of the meal is $35 and you want to leave 18% tip on the meal.

We want to find the total amount that will be paid.

First, we have to find 18% of $35 and then, add it to the original bill ($35).

18% of 35 is:

18/100 * 35 = $6.30

The tip is $6.30, therefore, the total amount paid will be:

$35 + $6.30 = $41.30

You will be paying $41.30 in total.

6 0

2 years ago

Find the missing angles for each problem, given that lines a and b are parallel (step by step)

marshall27 [118]

Answer:

1) 40°

2) 140°

3) 140°

5)40°

6)140°

7)140°

8)40°

Step-by-step explanation:

1 = 40 ° [Vertically Opposite angles are equal]

8 = 40 ° [Corresponding Angles are equal]

5 = 8 = 40° [Corresponding Angles are equal]

7 = 180° - 40° = 140° [Supplementary angles sum 180°]

6 = 140° [Vertically Opposite angles are equal]

2 = 140° [Corresponding Angles are equal]

3 = 140 ° [Vertically Opposite angles are equal]

3 0

2 years ago

Can someone help me out with this?? It’s due soon...

Inessa05 [86]

1. 1.25

2. 0.85

3. 0.2

Hope this helped.

5 0

2 years ago

Need help with number 3

marin [14]

IM THINKING B.)
HOPE ITS RIGHT LOL

3 0

3 years ago

Find the distance from P to l.

ki77a [65]

$Equation\ of\ a\ line\ from\ 2\ points (x_1;\ y_1);\ (x_2;\ y_2):\\\\y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\(-4;\ 2)\ and\ (3;-5)\\subtitute:\\\\y-2=\dfrac{-5-2}{3-(-4)}\cdot(x-(-4))\\\\y-2=\dfrac{-7}{7}\cdot(x+4)\\\\y-2=-(x+4)\\\\y-2=-x-4\ \ \ |add\ x\ and\ 4\ to\ both\ sides\\\\x+y+2=0\\-------------------\\$

$Point-line\ distance\\l:Ax+By+C=0;\ (x_0;\ y_0)\\\\d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}\\\\x+y+2=0\to A=1;\ B=1;\ C=2\\(1;\ 2)\to x_0=1;\ y_0=2\\subtitute\\\\d=\dfrac{|1\cdot1+1\cdot2+2|}{\sqrt{1^2+1^2}}=\dfrac{|1+2+2|}{\sqrt2}=\dfrac{|5|}{\sqrt2}=\dfrac{5}{\sqrt2}=\dfrac{5\cdot\sqrt2}{\sqrt2\cdot\sqrt2}\\\\=\boxed{\dfrac{5\sqrt2}{2}}$

4 0

3 years ago