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frutty [35]
3 years ago
9

A grocery store sells a bag of 5 lemons for $2.00 what is the unite cost of each lemon in the bag.

Mathematics
1 answer:
QveST [7]3 years ago
5 0

Answer:

.40 per lemon

Step-by-step explanation:

Take the total cost of the lemons and divide by the number of lemons

2.00 /5 = .40 per lemon

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PLEASE I NEED THIS TODAY!!!
Troyanec [42]

Answer:

Option D y\leq -3x-3 and y

Step-by-step explanation:

step 1

<u>Find the equation of the inequality A</u>

we know that

The solution of the inequality A (dashed line)  is the shaded area below the dashed line

The equation of the dashed line is y=-\frac{1}{2}x+2

therefore

The equation of the inequality A is y

step 2

<u>Find the equation of the inequality B</u>

we know that

The solution of the inequality B (solid line)  is the shaded area below the solid line

The equation of the solid line is y=-3x-3

therefore

The equation of the inequality B is y\leq -3x-3

The system of linear inequalities is

y\leq -3x-3 and y

6 0
4 years ago
Your buying candy for the next family get together. Lollipops cost $8 per bag . Chocolates cost $7 per bag. Smarties are $5 per
FinnZ [79.3K]
2 lollipops, 2 chocolates, and 4 smarties
4 0
3 years ago
What is (m+1)(m+1) multiplied
Zarrin [17]

Apply the "FOIL" method here. (m+1)(m+1) = m^2 + 1m + 1m + 1, or m^2 + 2m + 1.

4 0
3 years ago
A Niffler is a magical creature who loves to collect shiny objects and store them in their pouch, much like a kangaroo. The amou
Fudgin [204]

Answer:

Probability that a Niffler can hold more than 32 pounds of shiny objects in their pouch is 0.1515.

Step-by-step explanation:

We are given that the amount a Niffler can hold in their pouch is approximately normally distributed with a mean of 25 pounds of shiny objects and a standard deviation of 6.8 pounds.

Let X = <u><em>amount a Niffler can hold in their pouch</em></u>

So, X ~ Normal(\mu=25,\sigma^{2} =6.8^{2})

The z score probability distribution for normal distribution is given by;

                                Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean = 25 pounds

            \sigma = standard deviation = 6.8 pounds

Now, the probability that a Niffler can hold more than 32 pounds of shiny objects in their pouch is given by = P(X > 32 pounds)

  P(X > 32 pounds) = P( \frac{X-\mu}{\sigma} > \frac{32-25}{6.8} ) = P(Z > 1.03) = 1 - P(Z \leq 1.03)

                                                             = 1 - 0.8485 = 0.1515

<em>The above probability is calculated by looking at the value of x = 1.03 in the z table which has an area of 0.8485.</em>

<em />

Hence, the probability that a Niffler can hold more than 32 pounds of shiny objects in their pouch is 0.1515.

4 0
3 years ago
Is this correct ? check plsss...
d1i1m1o1n [39]
This looks one hundred percent correct to me and to check your answer you can plug in the answer you find to the first equation 


17=37+4f 
17=37+4(-5)
17=37-20
17=17 

great job!!
7 0
3 years ago
Read 2 more answers
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