All categories

3 years ago

Divide 216 into the ratio 1;5

Mathematics

1 answer:

SashulF [63]3 years ago

6 0

Answer:

1:5

Step-by-step explanation:

You might be interested in

Maria made blueberry muffins for a bake sale. Each muffin cost her about $0.25 to make. She spent $8.75 to make all of the muffi

Marat540 [252]

Answer: $17.50.

Step-by-step explanation;

If each muffin cost her $0.25 to make and she spent $8.75 making all muffins, then you can divide this amount to find out how many muffins she made (8.75/0.25), this is 35 muffins. If she sold each muffin for $0.50, multiply the number of muffins (35) by their price ($0.50), this is $17.50. Therefore, Maria received $17.50 from selling her muffins.

7 0

3 years ago

How to the stem and leaf plot and histogram differ? will mark brainliesttt

Verdich [7]

one has a number line with box and whiskers on top and a histogram is sort of like a bar chart it has bars going up that it the difference

hope this helps!

6 0

3 years ago

Read 2 more answers

A Finance professor created a multiple-choice examination with 100 questions for his students. Each question has five possible

seropon [69]

Answer:

a. 0.011 or 1.1%

b. 31.56% or 0.3156

c. 99.94% or 0.9994

d. 3.42% or 0.0342

Step-by-step explanation:

Given

Number of multiple choice questions = 100

Probability of success for students who have attended lectures and done their homework = 0.85

a. Using binomial distribution

Probability of correctly answering 90 or more questions out of 100

$= \sum^{100}_{x=90}\left {100} \atop {x}} \right C (0.85)^x(0.15)^{100-x}\\=0.011$

Since,

In Binomial Distribution

$P(X=x) =\left {h} \atop {x}} \right.C P^x q^{n-x}$

where $x=0,1,...,n$

and $q=1-P$

Probability is therefore 1.1% or 0.011

b. Probability of correctly answering 77 to 83 questions out of 100

$= \sum^{83}_{x=77}\left {100} \atop {x}} \right C (0.85)^x(0.15)^{100-x}\\=0.3156$

The probability is therefore 31.56% or 0.3156

c. Probability of correctly answering more than 73 questions out of 100

$= \sum^{100}_{x=73}\left {100} \atop {73}} \right C (0.85)^x(0.15)^{100-x}\\=0.9994$

The probability is therefore 99.94% or 0.9994

d. Assuming that the student has answered randomly

Probability of success = 1/5 = 0.2

Probability of failure = 1 - 0.2 = 0.8

Probability of answering 28 or more questions correctly

$= \sum^{100}_{x=28}\left {100} \atop {x}} \right.c (0.2)^x(0.8)^{100-x}\\=0.0342$

The probability is therefore 3.42%

5 0

3 years ago

How do I solve 3w+6=4w ?

Troyanec [42]

$3w+6=4w \ \ \ |\hbox{subtract 4w from both sides} \\
-w+6=0 \ \ \ |\hbox{subtract 6 from both sides} \\
-w=-6 \ \ \ |\hbox{mutiply both sides by (-1)} \\
w=6$

6 0

4 years ago

Read 2 more answers

Rationalise: (1) 4/(2+root3+root7) (2) 4/(2root3+root5)

Temka [501]

$\frac{4}{2+\sqrt3+\sqrt7}\cdot\frac{2-(\sqrt3+\sqrt7)}{2-(\sqrt3+\sqrt7)}=\frac{8-4\sqrt3-4\sqrt7}{2^2-(\sqrt3+\sqrt7)^2}=\frac{8-4\sqrt3-4\sqrt7}{4-3-2\sqrt{3\cdot7}-7}\\\\=\frac{8-4\sqrt3-4\sqrt7}{-6-2\sqrt{21}}=\frac{-2(2\sqrt3+2\sqrt7-4)}{-2(3+\sqrt{21})}=\frac{2\sqrt3+2\sqrt7-4}{3+\sqrt{21}}\cdot\frac{3-\sqrt{21}}{3-\sqrt{21}}\\\\=\frac{6\sqrt3-2\sqrt{63}+6\sqrt7-2\sqrt{147}-12+4\sqrt{21}}{3^2-(\sqrt{21})^2}=\frac{6\sqrt3-2\sqrt{9\cdot7}+6\sqrt7-2\sqrt{49\cdot3}-12+4\sqrt{21}}{9-21}$

$=\frac{6\sqrt3-6\sqrt7+6\sqrt7-14\sqrt3-12+4\sqrt{21}}{-12}=\frac{-8\sqrt3+4\sqrt{21}-12}{-12}=\frac{-4(2\sqrt3-\sqrt{21}+3)}{-12}\\\\=\frac{2\sqrt3-\sqrt{21}+3}{3}$

==============================================================

$\frac{4}{2\sqrt3+\sqrt5}\cdot\frac{2\sqrt3-\sqrt5}{2\sqrt3-\sqrt5}=\frac{8\sqrt3-4\sqrt5}{(2\sqrt3)^2-(\sqrt5)^2}=\frac{8\sqrt3-4\sqrt5}{4\cdot3-5}=\frac{8\sqrt3-4\sqrt5}{12-5}\\\\=\frac{8\sqrt3-4\sqrt5}{7}$

4 0

4 years ago

Read 2 more answers