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3 years ago

Which of the following could be used to evaluate the expression -6 ( 4 2/3) ?

Mathematics

2 answers:

const2013 [10]3 years ago

7 0

Answer:

A. $(-6)(4)+(-6)(\frac{2}{3})$

Step-by-step explanation:

Given expression,

$-6(4\frac{2}{3})$

Since, a mixed fraction can be written as the sum of whole number and fraction.

$\implies -6(4\frac{2}{3})=-6(4+\frac{2}{3})$

Using distributive property,

$-6(4\frac{2}{3})=(-6)(4)+(-6)(\frac{2}{3})$

Hence,

$(-6)(4)+(-6)(\frac{2}{3})$ is used to evaluate the expression.

Option 'A' is correct.

algol133 years ago

4 0

-6 (4 2/3)

-6 * (4 + 2/3)

-6 *4 + -6 * 2/3

Choice B

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3 years ago

Given: PS=RT, PQ=ST<br> Prove: QS=RS

ivanzaharov [21]

Answer:

I) Eq(1) reason: sum of segments of a straight line

II) Eq(2) reason: Given PQ = ST & PS = RT

III) Eq(3) reason: sum of segments of a straight line

IV) Eq(4) reason: Same value on right hand sides of eq(2) and eq(3) demands that we must equate their respective left hand sides

V) Eq(5) reason: Usage of collection of like terms and subtraction provided this equation.

Step-by-step explanation:

We are given that;

PS = RT and that PQ = ST

Now, we want to prove that QS = RS.

From the diagram, we can see that from concept of sum of segments of a straight line we can deduce that;

PQ + QS = PS - - - - (eq 1)

Now, from earlier we saw that PQ = ST & PS = RT

Thus putting ST for PQ & PS for RT in eq 1,we have;

ST + QS = RT - - - - (eq 2)

Again, from the line diagram, we can see that from concept of sum of segments of a straight line we can deduce that;

RS + ST = RT - - - - -(eq 3)

From eq(2) & eq(3) we can see that both left hand sides is equal to RT.

Thus, we can equate both left hand sides with each other to give;

ST + QS = RS + ST - - - (eq 4)

Subtracting ST from both sides gives;

ST - ST + QS = RS + ST - ST

This gives;

QS = RS - - - - (eq 5)

Thus;

QS = RS

Proved

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2 years ago

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3 years ago

Due to a manufacturing error, two cans of regular soda were accidentally filled with diet soda and placed into a 18-pack. Suppos

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Answer:

a) There is a 1.21% probability that both contain diet soda.

b) There is a 79.21% probability that both contain diet soda.

c) $P(X = 2)$ is unusual, $P(X = 0)$ is not unusual

d) There is a 19.58% probability that exactly one is diet and exactly one is regular.

Step-by-step explanation:

There are only two possible outcomes. Either the can has diet soda, or it hasn't. So we use the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

A number of sucesses $x$ is considered unusually low if $P(X \leq x) \leq 0.05$ and unusually high if $P(X \geq x) \geq 0.05$

In this problem, we have that:

Two cans are randomly chosen, so $n = 2$

Two out of 18 cans are filled with diet coke, so $\pi = \frac{2}{18} = 0.11$

a) Determine the probability that both contain diet soda. P(both diet soda)

That is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{2,2}(0.11)^{2}(0.89)^{0} = 0.0121$

There is a 1.21% probability that both contain diet soda.

b)Determine the probability that both contain regular soda. P(both regular)

That is P(X = 0).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{2,0}(0.11)^{0}(0.89)^{2} = 0.7921$

There is a 79.21% probability that both contain diet soda.

c) Would this be unusual?

We have that $P(X = 2)$ is unusual, since $P(X \geq 2) = P(X = 2) = 0.0121 \leq 0.05$

For P(X = 0), it is not unusually high nor unusually low.

d) Determine the probability that exactly one is diet and exactly one is regular. P(one diet and one regular)

That is P(X = 1).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{2,1}(0.11)^{1}(0.89)^{1} = 0.1958$

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