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1 year ago

Mateo has 7/2 pounds of apples. He needs 3/4 pounds of apples for each batch of applesauce. How many batches of applesauce can

Mathematics

1 answer:

Annette [7]1 year ago

6 0

Answer:

Step-by-step explanation:

Given that:

Total pounds of apple = 7/2 pounds

Pounds of apple needed per batch of applesauce = 3/4 pounds

Number of batches of applesauce which can be made :

Total pounds of apple / pounds of apple needed per batch

7/2 ÷ 3/4

7/2 * 4/3

= 28/6

= 4 4/6

= 4 2/3

Hence, number of batches of applesauce which can be made is 4

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Consider the equation 4 plus v equals 8 minus StartFraction 5 Over 3 EndFraction v.v + 4 + StartFraction 5 Over 3 EndFraction v

Anna [14]

Answer:

2v + 4 = 8

D on edg 2020

Step-by-step explanation:

9 0

2 years ago

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Pecans that cost $28.50 per kilogram were mixed with almonds that cost $22.25 per kilogram. How many kilograms of each were used

atroni [7]

Answer:

The pecans used in the mixture = 8 kg

The almonds used in the mixture = 17 kg

Step-by-step explanation:

The cost of per kg of pecans = $28.50

The cost of per kg of almonds = $22.25

Now, total weight of mixture = 25 kg

Let us assume the weight of pecans in the mixture = m kg

So, the amount of almonds in the mixture = Total weight - Weight of Pecans

= (25 - m) kg

Now, cost of m kg pecans = m x ( cost of 1 kg pecans) = m x ( $ 28.50)

= 28.50 m

cost of (25- m) kg almonds = (25 -m) x ( cost of 1 kg almonds)

= (25 - m) x ( $ 22.25)

= 556.25 - 22.25 m

Total cost of 25 kg of mixture = 25 x ( $24.25) = $606.25

Now, Total Cost of (almonds + Pecans) = Total cost of mixture

⇒28.50 m + 556.25 - 22.25 m = $606.25

or,6.25 m = 50

or, m = 50/6.25 = 8

Hence, the pecans used in the mixture = m = 8 kg

And the almonds used in the mixture = (25 - m) = 25 - 8 = 17 kg

5 0

1 year ago

1,250 an 15% sale an 6.5 sales tax

eimsori [14]

Given:
price = 1,250
sales discount = 15%
sales tax = 6.5%

The problem is unclear whether the price is the original price or the discounted price. I am assuming that the price is the original price.

Original price times sales discount rate is the value of sales discount
1,250 x 15% = 187.50

Original price less the value of sales discount is the discounted price.
1,250 - 187.50 = 1,062.50

Discounted price times sales tax rate is the value of the sales tax
1,062.50 x 6.5% = 69.06

Discounted price plus sales tax is the total cost of the purchase
1,062.50 + 69.06 = 1,131.56

4 0

2 years ago

A really bad carton of eggs contains spoiled eggs. An unsuspecting chef picks eggs at random for his ""Mega-Omelet Surprise."" F

Dima020 [189]

Answer:

(a) The probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b) The probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

Step-by-step explanation:

The complete question is:

A really bad carton of 18 eggs contains 8 spoiled eggs. An unsuspecting chef picks 5 eggs at random for his “Mega-Omelet Surprise.” Find the probability that the number of unspoiled eggs among the 5 selected is

(a) exactly 5

(b) 2 or fewer

Let X = number of unspoiled eggs in the bad carton of eggs.

Of the 18 eggs in the bad carton of eggs, 8 were spoiled eggs.

The probability of selecting an unspoiled egg is:

$P(X)=p=\frac{10}{18}=0.556$

A randomly selected egg is unspoiled or not is independent of the others.

It is provided that a chef picks 5 eggs at random.

The random variable X follows a Binomial distribution with parameters n = 5 and p = 0.556.

The success is defined as the selection of an unspoiled egg.

The probability mass function of X is given by:

$P(X=x)={5\choose x}(0.556)^{x}(1-0.556)^{5-x};\ x=0,1,2,3...$

(a)

Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:

$P(X=5)={5\choose 5}(0.556)^{5}(1-0.556)^{5-5}\\=1\times 0.05313\times 1\\=0.0531$

Thus, the probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b)

Compute the probability that of the 5 eggs selected 2 or less are unspoiled as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

$=\sum\imits^{2}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=0.0173+0.1080+0.2706\\=0.3959$

Thus, the probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c)

Compute the probability that of the 5 eggs selected more than 1 are unspoiled as follows:

P (X > 1) = 1 - P (X ≤ 1)

= 1 - P (X = 0) - P (X = 1)

$=1-\sum\limits^{1}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=1-0.0173-0.1080\\=0.8747$

Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

6 0

1 year ago

-3^(-x)-6=-3^x+10 Solve the equation below for x by graphing plz

vitfil [10]

To graph it, just graph $y=-3^{-x}-6$ and $y=-3^x+10$ and see where they intersect

I would like to solve it by using math and not graphing
if you don't want to see the math, just don't scroll down
the graphical meathod is above, first line, just read it

hmm
multiply both sides by -1
$3^{-x}+6=3^x-10$
multiply both sides by $3^x$
$3^0+6(3^x)=3^{2x}-10(3^x)$
$1+6(3^x)=3^{2x}-10(3^x)$
minus 1 from both sides and minus 6(3^x) from both sides
$0=3^{2x}-16(3^x)-1$
use u subsitution
$u=3^x$
we can rewrite it as
$0=u^2-16u-1$
now factor
I mean use quadratic formula
for $ax^2+bx+c=0$ $x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$
so for 0=u^2-16u-1, a=1, b=-16, c=-1
$u=\frac{-(-16)+/-\sqrt{(-16)^2-4(1)(-1)}{2(1)}$
$u=8+/-\sqrt{65}$
remember that u=3^x so u>0
if we have u=8+√65, it's fine, but u=8-√65 is negative and not allowed
so therfor
$u=8+\sqrt{65}=3^x$
$8+\sqrt{65}=3^x$
if you take the log base 3 of both sides you get
$log_3(8+\sqrt{65})=x$
if you use ln then
$ln(8+\sqrt{65})=xln(3)$
then
$\frac{ln(8+\sqrt{65})}{ln(3)}=x$

8 0

2 years ago

Read 2 more answers