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3 years ago

PLEASE HELP ASAP FOR 10 POINTS ❤️

Mathematics

1 answer:

olganol [36]3 years ago

8 0

Answer:

-y^4+4y

Step-by-step explanation:

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A cubic box is completely filled with 2,179 g of water. What is the length of one side of the box, in meters? m Explain your rea

Mazyrski [523]

Answer:

$l\approx 1.296\,cm$

Step-by-step explanation:

The water density is approximately 1 gram per cubic centimeter, which means that 1 gram of water occupies a volume of a cubic centimeter. Hence, 2.179 grams occupies a volume of 2.179 cubic centimeters.

Let assume volume occupied by the water has a cube-like shape, whose side has the following measure:

$l = \sqrt[3]{2.179\,m^{3}}$

$l\approx 1.296\,cm$

5 0

3 years ago

A line passes through the point (-8, 6) and has a slope of 3/2

Vladimir79 [104]

Answer:

y-6=3/2(x+8)

Step-by-step explanation:

so the point slope form is y-y₁=m(x-x₁)

so enter the points (-8, 6)

y-6= 3/2(x-(-8))

y-6=3/2(x+8)

3 0

3 years ago

Read 2 more answers

Pls help ASAPI don’t understand can you show steps too pls

tensa zangetsu [6.8K]

Answer:

x = -2 , y = -1

(-2,-1)

4 0

3 years ago

Can someone help me.

salantis [7]

Answer:

see explanation

Step-by-step explanation:

substitute the values of x in the table into g(x)

Using the rule of exponents

$a^{-m}$ = $\frac{1}{a^{m} }$

g(- 2) = $4^{-2}$ = $\frac{1}{4^{2} }$ = $\frac{1}{16}$

g(- 1) = $4^{-1}$ = $\frac{1}{4}$

g(0) = $4^{0}$ = 1

g(1) = $4^{1}$ = 4

g(2) = 4² = 16

8 0

3 years ago

Read 2 more answers

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