All categories

3 years ago

Help this might be easy for you not me

Mathematics

2 answers:

konstantin123 [22]3 years ago

8 0

The answer is $1.07 because 3.7 x 0.29 = 1.073 and 1.073 rounded to the next cent is $1.07

MariettaO [177]3 years ago

4 0

3.7 lbs, .29 per pound
.29 * 3.7 = C (cost)
in fact, .29p=C where c is cost and p is pounds of potatoes.
round it in terms of money so that it makes sense

You might be interested in

The system of equations relates the distance, y,

Monica [59]

Is it 7.1 or something I don’t know for sure

8 0

3 years ago

Can u guys PLEASE answer this question ASAP. Two wooden boxes, both with dimensions 80 cm, 1 m and 25 cm, are placed on the grou

Slav-nsk [51]

Answer:

34000cm or 340m

Step-by-step explanation:

We have to find total surface area.

dimensions of 1 wooden box =

l = 1m = 100cm

b = 80cm

h = 25cm

∴ In the case of 2 =

l = 100cm

b= 80cm

h = 25 + 25 = 50cm

T.S.A = 2(lb + lh + bh)

= 2(100*80 + 100*50+ 80*50)

= 2(8000+5000+4000)

= 2*17000

= 34000cm

= 340m

3 0

3 years ago

Solve the proportion for x.

Delvig [45]

$\frac{x}{8} = \frac{8}{64} \\ 64x = 8 \times 8 \\ 64x = 64 \\ x = \frac{64}{64} \\ \boxed{x = 1}$

Proportion for x is 1.
Use cross multiplication.

5 0

2 years ago

ANSWER ASAP!!! Find the length of the segment indicated

trapecia [35]

Answer:

14.45

Step-by-step explanation:

x^2+(7.1)^2=(16.1)^2

×^2+50.41=259.21

x^2=208.8

sqrt(x^2)=sqrt(208.8)

x=14.45

check.

14.45^2 + 7.1^2 = 16.1^2

4 0

3 years ago

Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,

MrRissso [65]

Answer:

$4.82\cdot 10^{-4}$

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

$p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769$

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

$n-1=51$

Therefore, the probability of drawing a three at the 2nd draw is

$p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784$

Then, at the third draw, the previous 2 cards are not replaced, so there are now

$n-2=50$

cards in the deck. So, the probability of drawing a jack is

$p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08$

Therefore, the total probability of drawing an ace, a three and then a jack is:

$p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}$

4 0

3 years ago