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finlep [7]
4 years ago
11

Someone help pleaseeee

Mathematics
1 answer:
qaws [65]4 years ago
7 0

Answer:

Part A = D

Part B = 5 and 50

Step-by-step explanation:

Part A:

area = 5*4x = 20x

so, 100 \leq 20x \leq  1000

Part B:

to leave it only by x, you divide both sides by 20

100/20 = 5, 1000/20 = 50

which 5\leq x\leq 50

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Sam chose one marble from a bag containing 2 red marbles, 3 white marbles, and 10 black marbles. Find the probability that the m
Anvisha [2.4K]

Answer:

P(black) = 2/3

Step-by-step explanation:

We need to find the total number of marbles

2 red marbles+ 3 white marbles+ 10 black marbles = 15 marbles


To find the probability that Sam chose a black marble, we take the number of black marbles and divide by the total number of marbles

P(black) = black marbles/total marbles

            = 10/15

Divide the top and bottom by 5

            = 2/3

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3 years ago
50 point question for this geometry test answer
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Step-by-step explanation:

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3 years ago
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About 543 out of every 1,000 people in United States owned a cell phone in 2003.In japan,the are was 68 for every 100 people.How
Softa [21]

Answer:

The greater percent of cell phone ownership in Japan than the U.S was <u>13.7%</u>.

Step-by-step explanation:

Given:

About 543 out of every 1,000 people in United States owned a cell phone in 2003.

In japan,the are was 68 for every 100 people.

Now, to find the greater percent of cell phone ownership in Japan than the U.S.

So, to get the percent of people owned cell phone in United States:

\frac{543}{1000}\times 100

=0.543\times 100

=54.3\%.

Now, to get the percent of people owned cell phone in Japan:

\frac{68}{100} \times 100\\\\=0.68\times 100\\\\=68\%.

So, to get the percent greater of cell phone ownership in Japan than the U.S we subtract the percent of people owned cell phone in United States from the percent of people owned cell phone in Japan:

68\%-54.3\%\\\\=13.7\%.

Therefore, the greater percent of cell phone ownership in Japan than the U.S was 13.7%.

5 0
3 years ago
1. Given points A(3, -5) and B(19, -1), find the coordinates of point C that sit 3/8 of the way along line AB, closer to A than
zaharov [31]

1. C(x, y) = (7.3, –3.9)

2. C(x, y) = (17, –1.5)

Solution:

Question 1:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction \frac{3}{8}.

Point divides segment in the ratio formula:

$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)

Here, x_1=3, y_1=-5, x_2=19, y_2=-1 and m = 3, n = 8

$C(x, y)=\left(\frac{3\times19+8\times3}{3+8} , \frac{3\times(-1)+8\times(-5)}{3+8}\right)

           $=\left(\frac{57+24}{11} , \frac{-3-40}{11}\right)

           $=\left(\frac{81}{11} , \frac{-43}{11}\right)

C(x, y) = (7.3, –3.9)

Question 2:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction \frac{3}{8}.

Point divides segment in the ratio formula:

$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)

Here, x_1=3, y_1=-5, x_2=19, y_2=-1 and m = 7, n = 1

$C(x, y)=\left(\frac{7\times19+1\times3}{7+1} , \frac{7\times(-1)+1\times(-5)}{7+1}\right)

           $=\left(\frac{133+3}{8} , \frac{-7-5}{8}\right)

           $=\left(\frac{136}{8} , \frac{-12}{8}\right)

C(x, y) = (17, –1.5)

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4 years ago
If tan (t) = -1.4 what is the tan(-t) ???
Svetradugi [14.3K]
1.4 is the answer pretty sure
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3 years ago
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