All categories

2 years ago

The cost of 10 oranges is $1.00. What is the cost of 5 dozen oranges?

2 answers:

5 0

They said 5 dozen. So 1 dozen is 12 12*5= 60 10 orange is $1 20 orange is $2 30 orange is $3 40 orange is $4 50 orange is $5 60 orange is $6 ( So the answer is $6) Hope it's help you easy way

Crank2 years ago

3 0

12 x 5 = 60 so The cost of 60 oranges is $6.00.

You might be interested in

What’s the answer??? Pls help asap this is due tonight

Katarina [22]

Answer:

Step-by-step explanation:

$5(3) - 3( - 5)$

$15 + 15$

$30$

hope it was helpful. any confusion u may ask

8 0

2 years ago

Read 2 more answers

Which algebraic expression represents “the quotient of a number and six”? 6 + w 6w 6 – w StartFraction w Over 6 EndFraction

S_A_V [24]

Answer:

The one with the fraction is the answer.

Step-by-step explanation:

Quotient means dividing, so the answer has to have a fraction sign in it. We can't see the answer options properly, so whatever one with the fraction is the one you should choose. I think it's the third option that you wrote.

Hope this helped!

8 0

3 years ago

Read 2 more answers

Given the probability histogram pictured for a discrete random variable X, what is μx?

zubka84 [21]

Answer:

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

If p=4 and q=10, evaluate the following expression:30/ q−p

Fittoniya [83]

..........the answer is 6

5 0

3 years ago

Read 2 more answers

(1 point) consider the function f(t)=⎧⎩⎨⎪⎪⎪⎪0,−5,−6,6,t<00≤t<11≤t<7t≥7;f(t)={0,t<0−5,0≤t<1−6,1≤t<76,t≥7; 1. wr

sergij07 [2.7K]

$f(t)=\begin{cases}0&\text{for }t$

Recall that

$u(t)=\begin{cases}0&\text{for }t$

Take it one piece at a time. For $t\ge0$ , we can scale $u(t)$ by -5:

$-5u(t)=\begin{cases}0&\text{for }t$

If we shift the argument by 1 and scale by -5, we have

$-5u(t-1)=\begin{cases}0&\text{for }t$

so if we subtract this from $-5u(t)$ , we'll end up with

$-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t$

For the next piece, we can add another scaled and shifted step like

$-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t$

so that

$-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t$

For the last piece, we add one more term:

$6u(t-7)=\begin{cases}0&\text{for }t$

and so putting everything together, we get $f(t)$ :

$f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)$
$f(t)\equiv-5u(t)-u(t-1)+12u(t-7)$

5 0

3 years ago