All categories

2 years ago

PLEASE HELPP!! DID I DO IT RIGHT???

Mathematics

1 answer:

astra-53 [7]2 years ago

8 0

Answer:

yea haha good job :-)

You might be interested in

I need help with these questions

Fofino [41]

Answer:

Rule up a graph sheet and text it to me so I can help

6 0

3 years ago

The number of households without a computer at home in 2009 was closest to A)801000 B)1153000 C)1737000 D)3433000 E)4017000

bagirrra123 [75]

Answer:

B) 1153000

Step-by-step explanation:

Given

See attachment

Required

Number of households without computer in 2009

From the attachment, we have:

$n=5170000$ --- survey

$p = 22.3\%$ --- proportion without computer

So, the actual number is:

$Number = p * n$

$Number = 22.3\% * 5170000$

$Number = 1152910$

$Number \approx 1153000$

<em>Hence: (B) is correct</em>

7 0

2 years ago

Aubrey’s monthly bank statement shows a total of $51 in fees for ATM withdrawals. If Audrey made 17 withdraws, what is Audrey ch

Tpy6a [65]

Answer:

Audrey was charged $3 for each withdrawal.

Step-by-step explanation:

We have to calculate the amount of money charged on one transaction.

In order to find the amount for one ATM withdrawal, the total amount for ATM withdrawals will be divided by number of withdrawals.

Given

$Total\ ATM\ withdrawal\ fees = \$51\\Total\ withdrawals = 17\\Amount\ for\ one\ withdrawal = \frac{Total\ fees}{No.\ of\ withdrawals}\\= \frac{51}{17}\\= \$3$

Hence,

Audrey was charged $3 for each withdrawal.

7 0

2 years ago

There are 26 third graders and 32 fourth graders going on a field trip. Each van can carry 10 students. How many cans are needed

iragen [17]

Answer:

26+32=10v

add numbers

58=10v

10v=58

divide both sides by 10

v=5.8

There will be 6 vans needed

58/6=9.6

There will be 10 students in four vans, and 9 students in two vans.

8 0

2 years ago

Limit definition for slope of the graph, equation of tangent line point for<br> f(x)=2x^2 at x=(-1)

Tems11 [23]

The slope of the tangent line to $f$ at $x=-1$ is given by the derivative of $f$ at that point:

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=\lim_{x\to-1}\frac{2x^2-2}{x+1}$

Factorize the numerator:

$2x^2-2=2(x^2-1)=2(x-1)(x+1)$

We have $x$ approaching -1; in particular, this means $x\neq-1$ , so that

$\dfrac{2x^2-2}{x+1}=\dfrac{2(x-1)(x+1)}{x+1}=2(x-1)$

Then

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{2x^2-2}{x+1}=\lim_{x\to-1}2(x-1)=2(-1-1)=-4$

and the tangent line's equation is

$y-f(-1)=f'(-1)(x-(-1))\implies y-4x-2$

6 0

2 years ago