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3 years ago

What is the sum of 2/5 and 1/3

Mathematics

2 answers:

VladimirAG [237]3 years ago

7 0

Answer:11/15

Step-by-step explanation:

Most common denominator is 15 multiply 2/5 by 3 and 1/3 by 5 and that’s what you get

worty [1.4K]3 years ago

6 0

The answer would be 11/15

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Divide 6 into 6 bowls to get 1

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2 years ago

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3 years ago

A randomized study compared the effectiveness of two mosquito repellents. The treatment group using Spray A reported a mean of 3

BlackZzzverrR [31]

Answer:

The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.

Step-by-step explanation:

Given that:

The treatment group using Spray A reported a mean of 3.5 mosquito bites.

The treatment group using Spray B reported a mean of 5.6 mosquito bites.

After the data are re-randomized; the differences of the means are shown in the dot plot. The dot plot is attached in the file below;

The best conclusion that can be make based on the data from the dot plot is :

The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.

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2 years ago

I don’t really get this question can someone help me out

Inessa [10]

angles EFH and GFH are congruents

for the second equivalence criterion two triangles are equivalents if they have respectively 2 angles and 1 side congruents

the two triangles have congruents the angle right angle (HGF and HEF) and the 55 degrees angles (EFH and GFH); then they have in common the HF side.

so the triangles are congruent and so all the sides are respectively congruent,for example EH with HG:

EH=HG

s+36=7s

s+36-7s=0

s-7s=-36

-6s=-36

6s=36

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sorry for my english

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3 years ago

A box of donuts containing 6 maple bars, 3 chocolate donuts, and 3 custard filled donuts is sitting on a counter in a work offic

Svetlanka [38]

Probabilities are used to determine the chances of selecting a kind of donut from the box.

The probability that Warren eats a chocolate donut, and then a custard filled donut is 0.068

The given parameters are:

$\mathbf{Bars = 6}$

$\mathbf{Chocolate = 3}$

$\mathbf{Custard= 3}$

The total number of donuts in the box is:

$\mathbf{Total= 6 + 3 + 3}$

$\mathbf{Total= 12}$

The probability of eating a chocolate donut, and then a custard filled donut is calculated using:

$\mathbf{Pr = \frac{Chocolate}{Total}\times \frac{Custard}{Total-1}}$

So, we have:

$\mathbf{Pr = \frac{3}{12}\times \frac{3}{12-1}}$

Simplify

$\mathbf{Pr = \frac{3}{12}\times \frac{3}{11}}$

Multiply

$\mathbf{Pr = \frac{9}{132}}$

Divide

$\mathbf{Pr = 0.068}$

Hence, the probability that Warren eats a chocolate donut, and then a custard filled donut is approximately 0.068