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3 years ago

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OK SO this is my report card, not my final. I have 2 more coming out so all three add up to my final. how much would i need in m

y classes to have an average of 90 smt for my FINAL report card?? pls send help

1 answer:

Hatshy [7]3 years ago

4 0

Answer:

Step-by-step explanation:

70% or more 70% is most likely a b-

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The slope of a line is 1, and the y-intercept is -1. what is the equation of the line written in slope-intercept form?

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The m in the equation is the slope and the y-intercept is the b.
y=1x-1
y=mx+b

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The volume of a cylinder of radius r and height h is given by V = π r 2 h. Then the height, in terms of the radius and volume, i

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3 years ago

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Indicate the equation of the given line in standard form. Show all your work for full credit. the line containing the median of

alukav5142 [94]

Answer:

* The equation of the median of the trapezoid is 10x + 6y = 39

Step-by-step explanation:

* Lets explain how to solve the problem

- The slope of the line whose end points are (x1 , y1) , (x2 , y2) is

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

- The mid point of the line whose end point are (x1 , y1) , (x2 , y2) is

$(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

- The standard form of the linear equation is Ax + BC = C, where

A , B , C are integers and A , B ≠ 0

- The median of a trapezoid is a segment that joins the midpoints of

the nonparallel sides

- It has two properties:

# It is parallel to both bases

# Its length equals half the sum of the base lengths

* Lets solve the problem

- The trapezoid has vertices R (-1 , 5) , S (! , 8) , T (7 , -2) , U (2 , 0)

- Lets find the slope of the 4 sides two find which of them are the

parallel bases and which of them are the non-parallel bases

# The side RS

∵ $m_{RS}=\frac{8-5}{1 - (-1)}=\frac{3}{2}$

# The side ST

∵ $m_{ST}=\frac{-2-8}{7-1}=\frac{-10}{6}=\frac{-5}{3}$

# The side TU

∵ $m_{TU}=\frac{0-(-2)}{2-7}=\frac{2}{-5}=\frac{-2}{5}$

# The side UR

∵ $m_{UR}=\frac{5-0}{-1-2}=\frac{5}{-3}=\frac{-5}{3}$

∵ The slope of ST = the slop UR

∴ ST// UR

∴ The parallel bases are ST and UR

∴ The nonparallel sides are RS and TU

- Lets find the midpoint of RS and TU to find the equation of the

median of the trapezoid

∵ The median of a trapezoid is a segment that joins the midpoints of

the nonparallel sides

∵ The midpoint of RS = $(\frac{-1+1}{2},\frac{5+8}{2})=(0,\frac{13}{2})$

∵ The median is parallel to both bases

∴ The slope of the median equal the slopes of the parallel bases = -5/3

∵ The form of the equation of a line is y = mx + c

∴ The equation of the median is y = -5/3 x + c

- To find c substitute x , y in the equation by the coordinates of the

midpoint of RS

∵ The mid point of Rs is (0 , 13/2)

∴ 13/2 = -5/3 (0) + c

∴ 13/2 = c

∴ The equation of the median is y = -5/3 x + 13/2

- Multiply the two sides by 6 to cancel the denominator

∴ The equation of the median is 6y = -10x + 39

- Add 10x to both sides

∴ The equation of the median is 10x + 6y = 39

* The equation of the median of the trapezoid is 10x + 6y = 39

7 0

3 years ago

John, sally, Natalie would all like to save some money. John decides that it would be best to save money in a jar in his closet

stiv31 [10]

Answer:

Part 1) John’s situation is modeled by a linear equation (see the explanation)

Part 2) $y=100x+300$

Part 3) $\$12,300$

Part 4) $\$2,700$

Part 5) Is a exponential growth function

Part 6) $A=6,000(1.07)^{t}$

Part 7) $\$11,802.91$

Part 8) $\$6,869.40$

Part 9) Is a exponential growth function

Part 10) $A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

Part 11) $\$13,591.41$

Part 12) $\$6,107.01$

Part 13) Natalie has the most money after 10 years

Part 14) Sally has the most money after 2 years

Step-by-step explanation:

Part 1) What type of equation models John’s situation?

Let

y ----> the total money saved in a jar

x ---> the time in months

The linear equation in slope intercept form

$y=mx+b$

The slope is equal to

$m=\$100\ per\ month$

The y-intercept or initial value is

$b=\$300$

so

$y=100x+300$

therefore

John’s situation is modeled by a linear equation

Part 2) Write the model equation for John’s situation

see part 1)

$y=100x+300$

Part 3) How much money will John have after 10 years?

Remember that

1 year is equal to 12 months

so

$10\ years=10(12)=120 months$

For x=120 months

substitute in the linear equation

$y=100(120)+300=\$12,300$

Part 4) How much money will John have after 2 years?

Remember that

1 year is equal to 12 months

so

$2\ years=2(12)=24\ months$

For x=24 months

substitute in the linear equation

$y=100(24)+300=\$2,700$

Part 5) What type of exponential model is Sally’s situation?

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$P=\$6,000\\ r=7\%=0.07\\n=1$

substitute in the formula above

$A=6,000(1+\frac{0.07}{1})^{1*t}\\ A=6,000(1.07)^{t}$

therefore

Is a exponential growth function

Part 6) Write the model equation for Sally’s situation

$A=6,000(1.07)^{t}$

see the Part 5)

Part 7) How much money will Sally have after 10 years?

For t=10 years

substitute the value of t in the exponential growth function

$A=6,000(1.07)^{10}=\$11,802.91$

Part 8) How much money will Sally have after 2 years?

For t=2 years

substitute the value of t in the exponential growth function

$A=6,000(1.07)^{2}=\$6,869.40$

Part 9) What type of exponential model is Natalie’s situation?

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$P=\$5,000\\r=10\%=0.10$

substitute in the formula above

$A=5,000(e)^{0.10t}$

Applying property of exponents

$A=5,000(1.1052)^{t}$

therefore

Is a exponential growth function

Part 10) Write the model equation for Natalie’s situation

$A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

see Part 9)

Part 11) How much money will Natalie have after 10 years?

For t=10 years

substitute

$A=5,000(e)^{0.10*10}=\$13,591.41$

Part 12) How much money will Natalie have after 2 years?

For t=2 years

substitute

$A=5,000(e)^{0.10*2}=\$6,107.01$

Part 13) Who will have the most money after 10 years?

Compare the final investment after 10 years of John, Sally, and Natalie

Natalie has the most money after 10 years

Part 14) Who will have the most money after 2 years?

Compare the final investment after 2 years of John, Sally, and Natalie

Sally has the most money after 2 years

3 0

3 years ago

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Adult panda weights are normally distributed with a mean of 200 pounds and a standard deviation of 20 pounds. The largest pandas

earnstyle [38]

$\mathbb P(X>250)=1-\mathbb P(X\le250)$

$\mathbb P(X\le250)=\mathbb P\left(\dfrac{X-200}{20}\le\dfrac{250-200}{20}\right)=\mathbb P(Z\le2.5)\approx0.9938$

which means

$\mathbb P(X>250)\approx1-0.9938=0.0062$

so approximately 0.62% of pandas weight over 250 pounds.

7 0

3 years ago

Read 2 more answers