Ask question

Ask question

All categories

3 years ago

12

Please help will give brainliest

2 answers:

lara31 [8.8K]3 years ago

5 0

Answer:

the answer is d

Step-by-step explanation:

-[-(7)]

-7= -(7)

-7= -7

Xelga [282]3 years ago

5 0

Answer:

a is the only statement that is true

Step-by-step explanation:

i hope this helps :)

You might be interested in

43,000 borrowed for 9 years with 7.5% interest if paid in full how much must be paid back

wlad13 [49]

assuming a simple interest rate

$\bf ~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$43000\\ r=rate\to 7.5\%\to \frac{7.5}{100}\dotfill &0.075\\ t=years\dotfill &9 \end{cases} \\\\\\ A=43000(1+0.075\cdot 9)\implies A=43000(1.675)\implies A=72025$

7 0

3 years ago

A taxi service charges a flat fee of $1.25 and $0.75 per mile. If Henri has $14.00, which of the following shows the number of m

maks197457 [2]

1.25 + 0.75x = 14
Solving for x gives the answer of 17. Thus, Henri can afford to travel 17 miles.

7 0

3 years ago

Robert wants to install windows in a bus barn to improve the airflow in the summer months. If Robert makes his own screens, he c

DedPeter [7]

=1/2

you add both on each side then subtract 2 from you anser

8 0

3 years ago

Y=-2/5x<br> a. direct <br> b. inverse <br> c. neither

disa [49]

Answer:

a. direct

Step-by-step explanation:

direct variation is y = kx and y = -2/5x is written in direct variation.

7 0

2 years ago

A deck of ordinary cards is shuffled and 13 cards are dealt. What is the probability that the last card dealt is an ace?

alexandr1967 [171]

Answer:

The probability that the last card dealt is an ace is $\frac{1}{13}$ .

Step-by-step explanation:

Given : A deck of ordinary cards is shuffled and 13 cards are dealt.

To find : What is the probability that the last card dealt is an ace?

Solution :

There are total 52 cards.

The total arrangement of cards is 52!.

There is 4 ace cards in total.

Arrangement for containing ace as the 13th card is $4\times 51!$ .

The probability that the last card dealt is an ace is

$P=\frac{4\times 51!}{52!}$

$P=\frac{4\times 51!}{52\times 51!}$

$P=\frac{4}{52}$

$P=\frac{1}{13}$

Therefore, the probability that the last card dealt is an ace is $\frac{1}{13}$ .

4 0

3 years ago