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3 years ago

(8+ 4)divided by 2 - 3

Mathematics

2 answers:

DiKsa [7]3 years ago

4 0

Answer:

Step-by-step explanation:

(8 + 4) / 2 - 3

12 / 2 -3

6 - 3

Tema [17]3 years ago

4 0

Answer:

Step-by-step explanation:

You do the parenthesis first (8+4)

Then you divide 12 by 2 if you are following Pemdas

6 - 3 = 3

Hope this helped!

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Find the domain and range of this

11Alexandr11 [23.1K]

Answer:

c on edg

Step-by-step explanation:

did same question

5 0

3 years ago

Log(2)4+log(2)3,16227766/log(2)20+log(2)8

Shalnov [3]

We know the rules of logarithms:
log a^b = b * log a
log a * b = log a + log b
and we know that : 10^0.5 = 3,166227766
So:
( log (2) 4 + log (2) 3,16227766) / ( log (2) 20 + log (2) 8 ) =
= ( 2 + log (2) 10^(0.5) ) / ( log (2) 2 + log (2) 10 + 3 ) =
= ( 2 + 0.5 log (2) 10 ) / ( 1 + log (2) 10 + 3 ) =
= ( 2 + 0.5 log (2) 10 / ( 4 + log (2) 10 ) =
= ( 2 + 0.5 log (2) 10 ) / 2 * ( 2 + 0.5 log (2) 10 ) = ( cancelling )
= 1 / 2 = 0.5

3 0

3 years ago

An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were

vladimir1956 [14]

Answer:

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

There is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units

Step-by-step explanation:

Step:-(1)

Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4

Mean of the first sample x₁⁻ =85

standard deviation of the first sample S₁ = 4

Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.

Mean of the first sample x₂⁻ =81

standard deviation of the first sample S₂ = 5

Step :-2

Null hypothesis: H₀: there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Alternative hypothesis :H₁: there is significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²

The test statistic

$Z= \frac{x_{1} -x_{2} }{\sqrt{\frac{S^2_{1} }{n_{1} } +\frac{S^2_{2} }{n_{2} } } }$

Given n₁=n₂=60.

$Z= \frac{85-81 }{\sqrt{\frac{(4)^2 }{60 } +\frac{5^2 }{60} } }$

On calculation, we get

Z = $\frac{4}{\sqrt{0.6833} }$

z = 4.8389

The tabulated value Z =1.96 at 0.05 level of significance.

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

Conclusion:-

there is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.

3 0

3 years ago

Multiply and simplify: (6x + 3y)(6x − 3y)

nasty-shy [4]

Answer:

36x² - 9y²

Step-by-step explanation:

We see that the given expression (6x + 3y)(6x - 3y) looks just like the expression (a + b)(a - b). When multiplied out, this is a difference of squares identity that I highly recommend you memorise:

(a + b)(a - b) = a² - b²

Here, a = 6x and b = 3y, so plug these in:

(6x + 3y)(6x - 3y) = (6x)² - (3y)² = 36x² - 9y²

5 0

3 years ago

Read 2 more answers

What is the value of x?

sladkih [1.3K]

If the angles are 90,45,45 so the sides except hypotenuse are congruent
and the hypotenuse is equal to :
${6 \sqrt{2} }^{2} + {6 \sqrt{2} }^{2} = {x}^{2} \\ {8.5}^{2} + {8.5}^{2} = {x}^{2} \\ 72.25 + 72.25 = {x}^{2} \\$
${x}^{2} = 144.5 \\ x = 12$

the answer is 12

hope this helps

6 0

4 years ago