All categories

3 years ago

Well give brainlist and free point.help me with this

Mathematics

2 answers:

Alika [10]3 years ago

8 0

The Pringle’s is 2.77$ per cup and the and the frito Lays are 2.82$ per cup so the pringle’s is less cost per cup

Korvikt [17]3 years ago

7 0

Answer:

Step-by-step explanation:

pringles 2.77

frito 2.82

mark this brainiest

You might be interested in

Pleasee helpp

max2010maxim [7]

Answer:

the answer is D because "difference" is a key word for subtraction

Step-by-step explanation:

8 0

3 years ago

-3 (-4) ×<br> 1 = –16<br> The answer babesss!!!!!

zvonat [6]

Answer:

x = -(4/3)

Step-by-step explanation:

$-3 (-4) \times1 = -16\\\\\mathrm{Divide\:both\:sides\:by\:}-3\left(-4\right)\times\:1\\\\\frac{-3\left(-4\right)x\times\:1}{-3\left(-4\right)\times\:1}=\frac{-16}{-3\left(-4\right)\times\:1}\\\\Simplify\\\\x=-\frac{4}{3}$

5 0

3 years ago

PLZ HELP ME ASAP!!!!!!!! THX!!!!!!!!!!!!!

topjm [15]

64/49

thats what I got but im not sure

5 0

3 years ago

Juanita borrowed $600 to purchase a new computer. She was charged 7% interest for two years. How much interest will Juanita pay?

motikmotik

I=prt. I=600(.07)(2). I=42(2). I=84. Juanita will pay $84 interest :)

4 0

3 years ago

Read 2 more answers

I roll a fair die twice and obtain two numbers X1= result of the first roll and X2= result of the second roll. Given that I know

azamat

By definition of conditional probability,

$P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{P((X_1=4\text{ or }X_2=4)\text{ and }X_1+X_2=7)}{P(X_1+X_2=7)}$

$=\dfrac{P((X_1=4\text{ and }X_1+X_2=7)\text{ or }(X_2=4\text{ and }X_1+X_2=7))}{P(X_1+X_2=7)}$

Assuming a standard 6-sided fair die,

if $X_1=4$ , then $X_1+X_2=7$ means $X_2=3$ ; otherwise,
if $X_2=4$ , then $X_1=3$ .

Both outcomes are mutually exclusive with probability $\frac1{36}$ each, hence total probability $\frac2{36}=\frac1{18}$ .

Of the 36 possible outcomes, there are 6 ways to sum the integers 1-6 to get 7:

(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

and so a sum of 7 occurs $\frac6{36}=\frac16$ of the time.

Then the probability we want is

$P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{\frac1{18}}{\frac16}=\frac13$

6 0

3 years ago