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2 years ago

How do I find the percent of budget?

Mathematics

1 answer:

ch4aika [34]2 years ago

6 0

Answer:

First, subtract the budgeted amount from the actual expense. If this expense was over budget, then the result will be positive.

Next, divide that number by the original budgeted amount and then multiply the result by 100 to get the percentage over budget. If your expenses were lower than your budgeted amount, then this number will be negative, describing the percentage under budget.

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S see Use the graph y=e^x to evaluate these expression e^0

nika2105 [10]

Answer:

$y= e^x$

And we want to evaluate e^0 using the graph

And as we can see in the plot the y intercept is the blue point with y=0 and that correspond with:

$y(0)= e^0 =1$

Step-by-step explanation:

For this problem we know the following function:

$y= e^x$

And we want to evaluate e^0 using the graph

And as we can see in the plot the y intercept is the blue point with y=0 and that correspond with:

$y(0)= e^0 =1$

5 0

2 years ago

Mark made a scale drawing of a soccer field, using a scale of .5 cm=1m. The actual length of the field is 110 m. What is the len

VikaD [51]

Answer:

The length of the field on the drawing is <u>55 cm</u> that is option <u>D</u>.

Step-by-step explanation:

Given:

Mark made a scale drawing of a soccer field, using a scale of .5 cm=1 m.

The actual length of the field is 110 m.

Now, to find the length of the field on the drawing.

Let the length of the field on drawing be $x\ cm.$

As given,

Drawing of a soccer field, using a scale of 0.5 cm=1 m.

So, 0.5 cm is equivalent to 1 m.

Thus, $x\ cm$ is equivalent to 110 m.

Now, to get the length of the field on the drawing by using cross multiplication:

$\frac{0.5}{1} =\frac{x}{110}$

By cross multiplying we get:

$55=x$

$x=55\ cm$

Therefore, the length of the field on the drawing is 55 cm that is option D.

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