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3 years ago

Complete the two-column proof.

2 answers:

8 0

88 - 6y = -1 substitution property (as 8 is substituted to replace x)
-6y = -89 subtraction property (88 is subtracted to the whole eqn)
y = 89/6 division property (-6 is the divisor)
89/6 = y transitive property (exchange of places results to the same value)

Nataliya [291]3 years ago

6 0

The correct answers are:

substitution property; subtraction property of equality; division property of equality; and reflexive property.

Explanation:

The difference between the given statement and the first statement is that 11x is replaced with 88. This is because we are told that x=8; 11(8) = 88. Substitution has been used here.

The difference between the first step and the second step is that 88 is no longer on the left hand side and the right hand side is -89. This is because to cancel 88, it has been subtracted from each side of the equation. This is the subtraction property of equality.

The difference between the second step and the third step is that -6 is no longer on the left side and the right hand side is 89/6. This is because to cancel -6 both sides were divided by -6. This is the division property of equality.

The difference between the third step and the last step is that the 89/6 and y have been flipped around the equals sign. This is the reflexive property.

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A man invests a total of $9,493in two savings accounts. One account yields 9% simple interest and the other 10% simple interest.

Tju [1.3M]

$ 1850 was invested in 9% account

Solution:

Given that

Total amount invested by man in two saving accounts = $9493

Simple interest on one account =9%

Simple interest on second account = 10%

Total interest earned = $930.80

Need to determine amount invested in 9 % account.

Let assume amount invested in account where Simple Interest is 9% = x

And assume amount invested in account where Simple Interest is 10% = y

As total amount invested in two accounts is $9493

=> x + y = 9493

=> y = 9493 - x ------(1)

$\text { Simple Interest }=\frac{\text { Amount Invested } \times \text {rate of interest } \times \text {time}}{100}$

$\begin{array}{l}{\text { Simple interest when rate of interest is } 9 \%=\frac{x \times 9 \times 1}{100}=\frac{9 x}{100}} \\\\ {\text { Simple interest when rate of interest is } 10 \%=\frac{y \times 10 \times 1}{100}=\frac{10 y}{100}}\end{array}$

As total interest earned = $930.80

$\begin{array}{l}{\Rightarrow \frac{9 x}{100}+\frac{10 y}{100}=930.80} \\\\ {\Rightarrow 9 x+10 y=930.80 \times 100} \\\\ {\Rightarrow 9 x+10 y=93080}\end{array}$

On substituting value of y from equation(1) in above equation , we get

9x + 10 (9493 – x) = 93080

=> 9x + 94930 – 10x = 93080

=> -x = 93080 – 94930

=> -x = -1850

=> x = 1850

Amount invested in account where Simple Interest is 9% = x = $1850

Hence $1850 was invested in 9% account.

5 0

3 years ago

Type a digit that makes this statement true. 4,330, 68 is divisible by 8.

telo118 [61]

Answer:

688 is divisible by 8

the digit I have typed is 8

Step-by-step explanation:

please mark me as brainlest

3 0

2 years ago

Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive

PtichkaEL [24]

Looks like the PMF is supposed to be

$\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}$

which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:

$\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65$

Next,

$\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25$

$\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5$

$\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2$
$\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5$
$\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}$

If $Y=X^2+1$ , then $X^2=Y-1\implies X=\sqrt{Y-1}$ , where we take the positive root because we know $X$ can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that $Y$ can take on the values $1^2+1=2$ , $2^2+1=5$ , and $5^2+1=26$ . At these values of $Y$ , we would have the same probability as we did for the respective value of $X$ . That is,

$\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}$

Part (5) is incomplete, so I'll stop here.

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