I need an answer asap

591,912 nearest thousand

Nana76 [90]

591,912 rounded to the nearest thousands = 592,000

4 0

3 years ago

Read 2 more answers

HELP ASAP! I will give brainliest answer

balu736 [363]

Answer: 1/3x + 2 = 1 answer.) x=-3

0.75x - 0.5x = -3 answer.) y = 17

Step-by-step explanation:

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3 0

3 years ago

Mr. Green teaches mathematics and his class recently finished a unit on statistics. The student scores on this unit are: 40 47 5

Harrizon [31]

Answer:

Mean = 64.46, Median = 62 and Mode = Bi-modal (50 and 62)

Range of the data is 55.

Step-by-step explanation:

We are given that Mr. Green teaches mathematics and his class recently finished a unit on statistics.

<u>The student scores on this unit are:</u> 40, 47, 50, 50, 50, 54, 56, 56, 60, 60, 62, 62, 62, 63, 65, 70, 70, 72, 76, 77, 80, 85, 85, 95.

We know that Measures of Central Tendency are: Mean, Median and Mode.

Mean is calculated as;

Mean = $\frac{\sum X}{n}$

where $\sum X$ = Sum of all values in the data

n = Number of observations = 24

So, Mean = $\frac{40+ 47+ 50+ 50+ 50+ 54+ 56+ 56+ 60 +60+ 62+ 62+ 62+ 63+ 65+ 70+ 70+ 72+ 76+ 77+ 80+ 85+ 85+ 95}{24}$

= $\frac{1547}{24}$ = 64.46

So, mean of data si 64.46.

For calculating Median, we have to observe that the number of observations (n) is even or odd, i.e.;

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.} }{2}$

Now here in our data, the number of observations is even, i.e. n = 24.

So, Median = $\frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.} }{2}$

= $\frac{(\frac{24}{2})^{th}\text{ obs.} +(\frac{24}{2}+1)^{th}\text{ obs.} }{2}$

= $\frac{(12)^{th}\text{ obs.} +(13)^{th}\text{ obs.} }{2}$

= $\frac{62 + 62 }{2}$ = $\frac{124}{2}$ = 62

Hence, the median of the data is 62.

A Mode is a value that appears maximum number of times in our data.

In our data, there are two values which appear maximum number of times, i.e. 50 and 62 as these both appear maximum 3 times in the data.

This means our data is Bi-modal with 50 and 62.

The Range is calculated as the difference between the highest and lowest value in the data.

Range = Highest value - Lowest value

= 95 - 40 = 55

Hence, range of the data is 55.

5 0

4 years ago

What is the volume of a cylinder with a diameter of 6 centimeters and height of 2 centimeters

Travka [436]

Answer:

56.55cm³

Step-by-step explanation:

Base radius: r = 7 m

Height: h = 11 m

3 0

3 years ago

The square of X varies inversely as the square root of Y and directly as the variable M. When M = 27 and Y = 16, then X = 9. Fin

irakobra [83]

Answer:

$Y = 441$

Step-by-step explanation:

Given

M = 27 when Y = 16 and X = 9

Required

Find Y when M = 7 and X = 2

We start by getting the algebraic representation of the given statement

$X^2 \alpha \frac{1}{\sqrt Y} \alpha M$

Convert the variation to an equation; we have

$X^2 = \frac{KM}{\sqrt Y}$

<em>Where K is the constant of variation;</em>

When M = 27; Y = 16; X = 9, the expression becomes

$9^2 = \frac{K * 27}{\sqrt{16}}$

This gives

$81 = \frac{k * 27}{4}$

Make K the subject of formula

$K = \frac{81* 4}{27}$

$K = \frac{324}{27}$

$K = 12$

Solving for Y when M = 7 and X = 2

Recall that $X^2 = \frac{KM}{\sqrt Y}$

Substitute values for K, M and X

$2^2 = \frac{12 * 7}{\sqrt{Y}}$

$4 = \frac{84}{\sqrt{Y}}$

Take square of both sides

$4^2 = (\frac{84}{\sqrt{Y}})^2$

$16 = \frac{7056}{Y}$

Make Y the subject of formula

$Y = \frac{7056}{16}$

$Y = 441$

3 0

3 years ago