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2 years ago

Who wants a free brainliest

Mathematics

2 answers:

andre [41]2 years ago

7 0

ME PLS I WOULD REALLY POSITIVELY LOVE THAT.

pickupchik [31]2 years ago

7 0

Me i do. please and thank you

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A standard toilet uses 1.6 gallons per flush. If a toilet is flushed five times per day, how much water will be used in a month

djyliett [7]

Answer:

I believe the answer is 48 gallons.

Step-by-step explanation:

Multiply 1.6 by 30 and you get 48

6 0

2 years ago

Read 2 more answers

Solve the equation for x.

PIT_PIT [208]

Get unknowns on one side and know on other side.

Divide both sides by -2
X= -7

4 0

2 years ago

Plz guys help i need asap<br> its part b im stuck

aleksklad [387]

Answer:

a.2nd quarter with 9 goals

b. 4.8 goals

c. 4 goals

Step-by-step explanation:

a. The mode is defined as the most appearing data point or the data point with the highest frequency..

From our data(for away goals):

1st quarter-2
2nd quarter-9
3rd quarter-7
4th quarter-4

Hence, the 2nd quarter has the mode for away goals with 9 goals.

b. Mean is defined as the average of a set of data points.

#We calculate the totals goals per quarter, sum over all quarters then divide by the number of games, 10:

$\bar x=\frac{1}{n}\sum{x_i}\\1^{st }_g=Away+Home=5+2=7\\\\2^{nd}_g=Away+Home=4+9=13\\\\3^{rd}_g=Away+Home=8+7=15\\\\4^{th}_g=Away+Home=9+4=13\\\\\bar x=\frac{1}{n}\sum{x_i}=\frac{1}{10}(7+13+15+13)=4.8$

Hence, the mean number of goals per quarter is 4.8 goals

c. To find the number of more home goals than away goals, we subtract from their summations as:

$g_m=\sum{g_h}-\sum{g_a}\\\\=(5+4+8+9)-(2+9+7+4)\\\\=26-22\\\\=4$

Hence, there are 4 more home goals than away goals.

4 0

2 years ago

The endpoints of (MP)are M(2,1) and P(12,6). If point K partitions (MP) in a ratio of MK:KP = 3:2, what are the coordinates of K

GREYUIT [131]

Answer:

K(8, 4)

Step-by-step explanation:

Given:

M(2, 1), P(12, 6)

MK:KP = 3:2

Required:

Coordinates of K

SOLUTION:

Coordinates of K can be determined using the formula below:

$x = \frac{mx_2 + nx_1}{m + n}$

$y = \frac{my_2 + ny_1}{m + n}$

Where,

$M(2, 1) = (x_1, y_1)$

$P(12, 6) = (x_2, y_2)$

$m = 3, n = 2$

Plug in the necessary values to find the coordinates of K:

$x = \frac{mx_2 + nx_1}{m + n}$

$x = \frac{3(12) + 2(2)}{3 + 2}$

$x = \frac{36 + 4}{5}$

$x = \frac{40}{5}$

$x = 8$

$y = \frac{my_2 + ny_1}{m + n}$

$y = \frac{3(6) + 2(1)}{3 + 2}$

$y = \frac{18 + 2}{5}$

$y = \frac{20}{5}$

$y = 4$

The coordinates of K = (8, 4)

3 0

3 years ago

: A new bear population that begins with 150 bears in 2000 decreases at a rate of 20% per year.

d1i1m1o1n [39]

Answer:

2 bears in 2020.

Step-by-step explanation:

We have been given that a new bear population that begins with 150 bears in 2000 decreases at a rate of 20% per year.

We will use exponential decay formula to solve our given problem as:

$y=a\cdot (1-r)^x$ , where,

y = Final quantity,

a = Initial value,

r = Decay rate in decimal form,

x = Time

$20\%=\frac{20}{100}=0.20$

Upon substituting our given values in above formula, we will get:

$y=150(1-0.20)^x$

$y=150(0.80)^x$ , where x corresponds to year 2000.

To find the population in 2020, we will substitute $x=20$ in our equation as:

$y=150(0.80)^{20}$

$y=150(0.011529215046)$

$y=1.7293822569$

$y\approx 2$

Therefore, 2 bears are there predicted to be in 2020.

Since population is decreasing so population is best described as exponential decay.

8 0

3 years ago