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nataly862011 [7]

2 years ago

13

(b) In a continuous series the average weight of some students is 75 kg and the sum of the weights is 6000 kg. Find the number o

f students.

1 answer:

Stolb23 [73]2 years ago

8 0

Answer:

the number of students is 80

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Alley bought 1 pound of potatoes costing $4 from a local grocery store last week. Her final bill included an additional sales ta

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$6.00 X 8% $6.48

this is the answer. hope this helps

8 0

2 years ago

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Show a way to count from 170 to 410 using tens and hundreds. circle at least 1 benchmark number

Anastasy [175]

Benchmark are numbers that are used as standards to which the rest of the data is compared to. When counting numbers using a number line, the benchmark numbers are the intervals written on the axis. For benchmark numbers of 10, the number line on top of the attached picture is shown. Starting from 170, the tick marks are added by 10, such that the next numbers are 180, 190, 200, and so on and so forth. When you want to find 410, just find the benchmark number 410.

The same applies to benchmark numbers in intervals of 100. If you want to find 170, used the benchmark numbers 100 and 200. Then, you estimate at which point represents 170. For 410, you base on the benchmark numbers 400 and 500.

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2 years ago

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Russell has a collection of 1,200 pennies. Of these pennies, 25% are dated before 1980, 35% are dated from 1980 to 2000, and the

STatiana [176]

480 pennies.

25% are dated before 1980, therefore:
1200(total):100 = 12 (1%)
12(1%)x25 = 300 (25%)

35% are dated from 1980 to 2000, therefore:
12(1%)x35 = 420 (35%)

25% are dated before 1980, 35% are dated from 1980 to 2000 and the rest are dated after 2000.
25% + 35% = 60%.
300(25%) + 420(35%) = 720.
60% = 720.
100%-60% = 40%.
40% is the amount of pennies that are dated after 2000.

The total (100%) of the pennies is 1200.
We also know that 60% of the total (1200) is 720.

1200(100%) - 720(60%) = 480(40%).

Therefore, the amount of pennies dated after 2000 in Russell’s Collection is 480.

7 0

2 years ago

In a race, Car #1 went an average of 165 miles per hour for x hours before dropping out. Car #2 went an average of 175 miles per

Akimi4 [234]

When you think about it. It's actually quite easy.
The number are the amount of miles, and the variables are the amount of hours. So we would just have to multiply them. So, by putting the number together we would be able to multiply them once we have the numbers for the variables. Since we dont have them, we can just make the inequality. It says that car 1 traveled at LEAST that of car 2. Showing that car 1 may have traveled farther. SOOOOO........ the inequality should look like this:

165x <span>≥ 175y</span>

7 0

2 years ago

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The sum of the measures of the angles of a triangle is 180. The sum of the measures of the second and third angles is five times

rusak2 [61]

Answer:

$x=30\\y=68\\z=82$

Step-by-step explanation:

x = measure of the first angle

y = measure of the second angle

z = measure of the third angle

The sum of the measures of the second and third (y+z) is five times the measure of the first angle (=5x)

$y+z=5x$

The third angle is 14 more than the second

$z=y+14$

And remember that the sum of these three angles must be equal to 180.

$x+y+z=180$

Let's take these equations

$y+z=5x\\z=y+14\\x+y+z=180$

If you take a look at the first equation, we have y+z = 5x and we have y+z in the third equation as well, we can replace that....

$x+y+z=180\\x+(y+z)=180\\x+(5x)=180$

Distribute the + sign

$x+5x=180$

Combine like terms;

$6x=180$

Divide by 6.

$x=\frac{180}{6}\\ x=30$

We have now defined that the measure of the first angle is 30º.

Let's take another equation... for example $z=y+14$

I'm going to take this one because if I replace x and z in the third equation, all I'll have left will be y.

$x+y+z=180\\30+y+(y+14)=180$

Distribute the + sign and Combine like terms;

$30+y+y+14=180\\44+2y=180\\$

Subtract 44 to isolate 2y.

$2y=180-44\\2y=136$

Now divide by 2.

$y=\frac{136}{2}\\ y=68$

We already have the value of x and y. Once again, replacing this in the third equation will leave us with z to solve for.

$x+y+z=180\\30+68+z=180\\98+z=180\\z=180-98\\z=82$

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2 years ago