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2 years ago

WHAT DOES B EQUAL I NEED HELP PLEASEEE

Mathematics

2 answers:

Fittoniya [83]2 years ago

6 0

Answer:

Step-by-step explanation:

hope this helps

Kipish [7]2 years ago

3 0

Answer:

b is 0

Step-by-step explanation:

b is just 0

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When I have walked 20% of the way to school, I have 1200 metres more to walk than when I have 20% of the walk remaining.

Orlov [11]

Given:

I have walked 20% of the way to school.

I have 1200 metres more to walk than when I have 20% of the walk remaining.

To find:

The distance from home to school.

Solution:

Let x be the distance from home to school.

I have already walked 20% of the way to school and i have 1200 metres more to walk than when I have 20% of the walk remaining.

It means 1200 is $100\%-20\%-20\%=60\%$ of the total distance from home to school.

$1200=\dfrac{60}{100}x$

$1200=0.6x$

$\dfrac{1200}{0.6}=x$

$2000=x$

Therefore, the distance from home to school is 2000 metres.

7 0

2 years ago

suppose the linear regression line y= 2.4x + 3.1 predicts the time,in minutes it takes you to be served at a food stand if there

Scorpion4ik [409]

Answer:

Just substitute "x" with "3".

Step-by-step explanation:

y = 2.4x + 3.1

y = 2.4(3) +3.1

y = 7.2 + 3.1

y= 10.3

Answer: 10.3 minutes

3 0

3 years ago

Read 2 more answers

Sheza and Hadia had 12579 paper stars. Sheza gave 2214 paper stars to Hadia. Now Hadia had twice as many paper stars as Sheza. H

makvit [3.9K]

Answer:

10365

Step-by-step explanation:

12579 - 2214

= 10365

7 0

2 years ago

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

Part of the area under this ramp serves for storage. Items can be stored only in the unshaded part of the ramp

Mekhanik [1.2K]

Answer:

48.1 m³ available, 2.5m³ not available

Step-by-step explanation:

Tan(X) = 4.5/(5.8+1.7)

Tan X = 3/5

Hypotenuse² = 3² + 5² = 34

Sin X = 3/sqrt(34)

X = 30.96

1/sinX = 1.7/SinY

Y = 61.003°

Third angle: 180 - X - Y = 88.037

Volume

= ½ × 1.7 × 1 × sin(88.038) × 3

= .

= 2.548

= 2.5m³

Available for storage:

½×4.5×7.5×3 - 2.548

= 48.077 = 48.1 m³

7 0

2 years ago