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2 years ago

Please help it’s a test

Mathematics

2 answers:

gogolik [260]2 years ago

8 0

Answer:

x =7

Step-by-step explanation:

60/3 =20

20 = 2x +6

14 = 2x

14/2 = x

7 =x

x = 7

CHeck Back

2 x 7 + 6 = 20

20 x 3 = 60

I am Lyosha [343]2 years ago

7 0

Answer:

x=7

Step-by-step explanation:

3(2x+6)=60

2x+6=20

x+3=10

x=7

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PLEASE HELP QUICK, I need to know the answer.

Marysya12 [62]

The equivalent expression is $7^{2}$ .

We have the following expression -

$\frac{7^{-3} }{7^{-5} }$

We have to find its equivalent value.

<h3>What is the equivalent expression of the following expression-</h3>

$f(x, y) =\frac{A^{x} }{A^{y} }$

In order to divide two exponents with same base, we use the 'Quotient property of exponents.

The equivalent expression can be written as -

f(x, y) = $A^{x} A^{-y} = A^{x-y}$

In the question given we have -

$\frac{7^{-3} }{7^{-5} }$

Using the property discussed above -

$7^{-3}\;7^{5}$ = $7^{-3+5}$ = $7^{2}$

Hence, the equivalent expression is $7^{2}$ .

To solve more questions on exponents, visit the link below -

brainly.com/question/17173276

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3 0

2 years ago

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an artificial lake is in the shape of a rectangle and has an area of 9/20 square mile. the width of the lake is 1/5 the length o

Gnom [1K]

X=length
y=width
Area (rectangle)=length x width
we suggest this system of equations:
y=x/5
xy=9/20
solve this system of equations by substitution method:
x(x/5)=9/20
x²/5=9/20
Least common multiple=20
4x²=9
x²=9/4
x=⁺₋√(9/4)
we have two solutions:
x₁=-3/2 it does not validate
x₂=3/2 ⇒y=x/5=(3/2)/5=3/10

The dimensions of the lake are:
lengh=3/2 miles
width=3/10 miles

to check:
Area=3/2 miles x 3/10 miles=9/20 miles².
width is 1/5 the length of the lake ⇒ 3/10 miles=(3/2) /5 miles

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3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

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3 years ago

Write 1 2/3 as the sum of a whole number and two fractions that have the same denominator.

larisa [96]

1+1/3+1/3 because 1/3 and 1/3 have the same denominator.

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2 years ago

Q1.Simplify:

solniwko [45]

$\textbf{(i)}\\\\\{1^3 +2^3 \} \times \left(\dfrac 13 \right)^2\\\\=(1+8) \left(\dfrac 19 \right)\\\\=9\left(\dfrac 19 \right)\\\\=1\\\\$

$\textbf{(ii)}\\\\\{5^{-1}\times 4^{-1}\}^2\\\\=\left(5^{-1} \right)^2 \times \left(4^{-1} \right)^2~~~~~~~~~~~;[(ab)^m = a^mb^m]\\\\=5^{-2}\times 4^{-2}~~~~~~~~~~~~~~~~~~~~;[(a^m)^n = a^{mn}]\\\\=\dfrac 1{5^2} \times \dfrac 1{4^2}~~~~~~~~~~~~~~~~~~~~~~;\left[a^{-m} = \dfrac 1{a^m},~ a\neq 0 \right]\\\\=\dfrac{1}{400}\\\\=0.0025\\\\$

$\textbf{(iii)}\\\\\left\{ \left(\dfrac 13 \right)^{-2} \times \left( \dfrac 12 \right)^{-2}\right\}\div \left(\dfrac 14 \right)^{-3}\\\\\\=\left( \dfrac 13 \times \dfrac 12 \right)^{-2} \div\left(4^{-1}\right)^{-3}\\\\\\=\left(\dfrac 16 \right)^{-2} \div 4^3\\\\\\=\left(6^{-1} \right)^{-2}\div 4^3\\\\\\=6^2 \div 4^3\\\\\\=36\div 64\\\\\\=\dfrac{9}{16}\\\\\\=0.5625$

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2 years ago

Read 2 more answers