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3 years ago

Jasmine drew this display to represent the amount of

Mathematics

2 answers:

Margarita [4]3 years ago

7 0

Answer:

The first and third one is correct.

Give the person above brainliest

ED2021

Molodets [167]3 years ago

5 0

Answer:

its one and three on edge

Step-by-step explanation:

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What is the produce? -9x(5-2x)

Marta_Voda [28]

Answer:

279x

Step-by-step explanation:

5 0

3 years ago

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A sphere has a surface area of about 999 square millimeters. What is the radius, in millimeters, of the sphere

motikmotik

Given that, a sphere has a surface area of about 999 square millimeters.

Formula to find the surface area of sphere is,

S= 4πr² Where r = radius of the sphere and S= surface area.

Given S= 999 square millimeters.

So, we can write:

4πr² = 999

4 * 3.14 * r² = 999 Since, π = 3.14

12.56 *r² = 999

$r^2=\frac{999}{12.56}$ Divide each sides by 12.56.

r² = 79.53821656

r = √79.53821656 Taking square root to each sides of equation.

r = 8.918420071

So, r = 8.918 Rounded to nearest thousandth.

So, radius of the sphere is 8.918 millimeters.

Hope this helps you!

5 0

3 years ago

1,873 is roundes to the nearest thousand is

Serjik [45]

It would be 2,000. 1,873 rounded to the next number is 2,000, if the number was 1,499 you would round down but since the second number is higher than 5 you round up no matter what

5 0

3 years ago

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A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use of one particular cust

Fiesta28 [93]

SOLUTION

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given set of values

$321,397,559,454,475,324,482,558,369,513,385,360,459,403,498,477,361,366,372,320$

STEP 2: Write the formula for calculating the Standard deviation of a set of numbers

$\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ where\text{ }x_i\text{ are data points,} \\ \bar{x}\text{ is the mean} \\ \text{n is the number of values in the data set} \end{gathered}$

STEP 3: Calculate the mean

$\begin{gathered} \bar{x}=\frac{\sum ^{}_{}x_i}{n} \\ \bar{x}=\frac{\sum ^{}_{}(321,397,559,454,475,324,482,558,369,513,385,360,459,403,498,477,361,366,372,320)}{20} \\ \bar{x}=\frac{8453}{20}=422.65 \end{gathered}$

STEP 4: Calculate the Standard deviation

$\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ \sum ^{}_{}(x_i-\bar{x})^2\Rightarrow\text{Sum of squares of differences} \\ \Rightarrow10332.7225+657.9225+18591.3225+982.8225+2740.52251+9731.8225+3522.4225+18319.6225+2878.3225 \\ +8163.1225+1417.5225+3925.0225+1321.3225+386.1225+5677.6225+2953.9225+3800.7225 \\ +3209.2225+2565.4225+10537.0225 \\ \text{Sum}\Rightarrow108974.0275 \\ \\ S\tan dard\text{ deviation}=\sqrt[]{\frac{111714.55}{20-1}}=\sqrt[]{\frac{111714.55}{19}} \\ \Rightarrow\sqrt[]{5879.713158}=76.67928767 \\ \\ S\tan dard\text{ deviation}\approx76.68 \end{gathered}$

Hence, the standard deviation of the given set of numbers is approximately 76.68 to 2 decimal places.

STEP 5: Calculate the First and third quartile

$\begin{gathered} \text{IQR}=Q_3-Q_1 \\ \\ To\text{ get }Q_1 \\ We\text{ first arrange the data in ascending order} \\ \mathrm{Arrange\: the\: terms\: in\: ascending\: order} \\ 320,\: 321,\: 324,\: 360,\: 361,\: 366,\: 369,\: 372,\: 385,\: 397,\: 403,\: 454,\: 459,\: 475,\: 477,\: 482,\: 498,\: 513,\: 558,\: 559 \\ Q_1=(\frac{n+1}{4})th \\ Q_1=(\frac{20+1}{4})th=\frac{21}{4}th=5.25th\Rightarrow\frac{361+366}{2}=\frac{727}{2}=363.5 \\ \\ To\text{ get }Q_3 \\ Q_3=(\frac{3(n+1)}{4})th=\frac{3\times21}{4}=\frac{63}{4}=15.75th\Rightarrow\frac{477+482}{2}=\frac{959}{2}=479.5 \end{gathered}$

STEP 6: Find the Interquartile Range

$\begin{gathered} IQR=Q_3-Q_1 \\ \text{IQR}=479.5-363.5 \\ \text{IQR}=116 \end{gathered}$

Hence, the interquartile range of the data is 116

3 0

1 year ago

Slope of 2y = -x + 10

777dan777 [17]

The slope would be -1 because of the x

6 0

3 years ago

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