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3 years ago

12qt=_______gal ‍ ‍ ‍ ‍ ‍ ‍ ‍ ‍ ‍ ‍ ‍ ‍ ‍ ‍

Mathematics

2 answers:

coldgirl [10]3 years ago

5 0

Answer:

12qt = 3gal

Step-by-step explanation:

Anna007 [38]3 years ago

3 0

12 quarts is 3 gallons

Solve: 12 quarts = ? gallon

Let ? repersent x.

Solution: (Nothing to really explain since this is like once you see it you answer it there's no answer pattern or a step by step explanation. Good Luck on other problems like this.

1 quart = 0.25 gallon

2 quart = 0.5 gallon

5 quart = 1.25

10 quart = 2.5

As we know all this stuff, we can add 1, 2, 5, or 10 quarts to make a sum of 12.

Here are some equations

$2 + 2+2+2+2+2$

$10+2$

There is way more but these ones are the easiest. Yes, there is $1+1$ all the way to 12, but we don't need it.

Now let's calculate the first one 2 + alot of 2's. To do that:

1. Change all the 2's into 0.5.

$0.5 + 0.5 +0.5 + 0.5 + 0.5 + 0.5 = 0.5*6$

2. Multiply

$0.5*6$ is 3.

Therefore, 12 quarts = 3 gallons.

Now, let's try the 10+2 one. To do that"

1. Change 10 into 2.5 and 2 into 0.5

$2.5+0.5$

2. Add

$2.5+0.5=3$

Therefore, 12 quarts = 3 gallons.

Hope this helps!

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A regular polygon with an interior angle sum of 9000°? If so, what is it?

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The question ask to determine what is a regular polygon that contains an interior angle of 9000 degree, base on my research, according to the geometrical research, I would say that the said polygon is a 52 sided convex polygon. I hope this would help

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3 years ago

Which number is a factor of 12 but not a multiple of 3?

Delvig [45]

First list all the factors of 12;

1, 2, 3, 4, 6, 12

now let's see which ones are multiples of 3;

3, 6, 12

so all we have left are numbers 1, 2 and 4.

hope that helps, God bless!

6 0

3 years ago

Given: A, B, and C What is the value of X in the matrix equation AX + B = C?

Ksju [112]

Answer:

Option (2)

Step-by-step explanation:

Given expression is, AX + B = C

$A=\begin{bmatrix}-3 & -4\\ 1 & 0\end{bmatrix}$

$B=\begin{bmatrix}-7 & -9\\ 4 & -1\end{bmatrix}$

$C=\begin{bmatrix}-42 & -20\\ 5 & 4\end{bmatrix}$

AX + B = C

AX = C - B

C - B = $\begin{bmatrix}-42 & -20\\ 5 & 4\end{bmatrix}-\begin{bmatrix}-7 & -9\\ 4 & -1\end{bmatrix}$ = $\begin{bmatrix}-42+7 & -20+9\\ 5-4 & 4+1\end{bmatrix}$

C - B = $\begin{bmatrix}-35 & -11\\ 1 & 5\end{bmatrix}$

Let $X=\begin{bmatrix}a & b\\ c & d\end{bmatrix}$

AX = $\begin{bmatrix}-3 & -4\\ 1 & 0\end{bmatrix}\times \begin{bmatrix}a & b\\ c & d\end{bmatrix}$

= $\begin{bmatrix}(-3a-4c) & (-3b-4d)\\ a & b\end{bmatrix}$

Since AX = C - B

$\begin{bmatrix}(-3a-4c) & (-3b-4d)\\ a & b\end{bmatrix}=\begin{bmatrix}-35 & -11\\ 1 & 5\end{bmatrix}$

Therefore, a = 1, b = 5

(-3a - 4c) = -35

3(1) + 4c = 35

3 + 4c = 35

4c = 32

c = 8

And (-3b - 4d) = -11

3(5) + 4d = 11

4d = -4

d = -1

Therefore, Option (2). X = $\begin{bmatrix}1 & 5\\ 8 & -1\end{bmatrix}$ will be the answer.

7 0

3 years ago

A study sampled 350 upperclassmen (Group 1) and 250 underclassmen (Group 2) at high schools around the city of Houston. The stud

iris [78.8K]

Answer:

$-0.048$

Do not reject $H_0:P_1-P_2=0$

Step-by-step explanation:

From the question we are told that

Sample size $n_1=350$

Sample size $n_2=250$

Sample proportion 1 $\hat P= \frac{25}{350} =>0.07$

Sample proportion 2 $\hat P= \frac{19}{250} =>0.076$

95% confidence interval

Generally for 95% confidence level

Level of significance

$\alpha = 1-0.95=>0.05$

$\alpha /2=\frac{0.05}{2} =>0.025$

Therefore

$Z_a_/_2=1.96$

Generally the equation for confidence interval between $P_1 - P_2$ is mathematically given as

$(\hat P_1-\hat P_2)\pm Z_a_/_2\sqrt{\frac{\hat P_1(1-\hat P_1)}{n_1}+\frac{\hat P_2(1-\hat P_2)}{n_2} }$

$(0.07-0.076)\pm 1.96\sqrt{\frac{0.07(1-0.07)}{350}+\frac{0.076(1-0.076)}{250} }$

$(0.07-0.076)\pm 1.96\sqrt{4.66896*10^-^4 }$

$(-0.006)\pm 0.042$

$(-0.006)- 0.042=>-0.048$

$(-0.006)+ 0.042=>0.036$

Therefore

Confidence interval is

$-0.048$

Conclusion

Given the confidence interval has zero

Therefore do not reject $H_0:P_1-P_2=0$

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2 years ago

A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th

Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

The bearing of the buoy from the is approximately 307.35°

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

$\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}$

Therefore;

$\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}$

Which gives;

$\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}$

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}$

Similarly, we can get;

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}$

The angle subtended by the distance of the buoy from the port, C is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ 307.35°

Learn more about bearing in mathematics here:

brainly.com/question/23427938

5 0

2 years ago