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3 years ago

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Which pair of expressions is equivalent using the Associative Property of Multiplication? 6(4a ⋅ 2) = 24a ⋅ 12 6(4a ⋅ 2) = (4a ⋅

2) ⋅ 6 6(4a ⋅ 2) = 6 ⋅ 4a ⋅ 2 6(4a ⋅ 2) = (6 ⋅ 4a) ⋅ 2

1 answer:

galben [10]3 years ago

4 0

<span>Associative Property of Multiplication : (ab)c = a(bc)
</span>so
answer is

6(4a ⋅ 2) = (6 ⋅ 4a) ⋅ 2

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PLEASE HELP ME! Find the cube roots of 125(cos 288° + i sin 288°).

shutvik [7]

◆ COMPLEX NUMBERS ◆

125 ( cos 288 + i sin 288 ) can be written as -

125.e^i( 288)
125.e^i( 288 +360 )
125.e^i( 288+ 720)

[ As , multiples of 360 can be added to an angle without changing any trigonometric functions or sign ]

To find the cube root , take the cube root of above 3 expressions ,

We get -
5 e^( i 96 )
5 e^( i 216 )
5 e^( i 336 )

Now using Euler's formula , We rewrite above as -

5 ( cos 96 + i sin 96 )
5(c os 216 + i sin 216 )
5 ( cos 336 + i sin 336 ) Ans.

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3 years ago

Patsy and Tyra together have $50. When Patsy gifts Tyra $12, they both have the same amount. How much did they each have in the

Radda [10]

x=x-12

x+x-12=50

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3 years ago

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have

$P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453$

The sum of the probabilities is

$P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981$

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3 years ago

PLEASE ANSWER ASAP! YOUR ANSWER MUST REQUIRE AN EXPLANATION IN ORDER TO RECEIVE 10 POINTS AND THE BRAINLIEST ANSWER!!! THANKS!!!

KIM [24]

Hello,

Using vectors and scalar product:

$[(a+c)*\vec{i}+(b-0)*\vec{j} ].[(a-c)*\vec{i}+(b-0)*\vec{j}]=0$
Thus
$\dfrac{b^2}{(a+c)(a-c)}=1$

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Answer A

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2 years ago

PLZ ANSWER ASASP ALMOST DUE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Burka [1]

Answer:

D) The third side could be 4 meters long.

Step-by-step explanation:

The third side of a triangle must be greater than the difference and less than the sum of the other two sides.

5-4=1

5+4=9

So the third side must be GREATER THAN 1 and LESS THAN 9. (1<x<9)

Therefore, the answer can't be A or C. The second one <em>could </em>be correct, depending on how much less than 4 it is. But the last one is definitely correct.

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2 years ago