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a_sh-v [17]
3 years ago
13

-2/3x + 9 = 4/3x -3​

Mathematics
2 answers:
olchik [2.2K]3 years ago
7 0

Answer:

x = 6

Step-by-step explanation:

To solve, move all the x’s to one side and all the numbers. Remember when moving numbers to the other side of an equation, the sign changes from either a positive to a negative or vise versa.

9 + 3 = 4/3x + 2/3x

Next, add the like terms.

9+3=12

When adding fractions, add the numerator (top number), but first change denominator (bottom number) if not the same. Here, we don’t need to.

4/3x+2/3x=6/3x

So,

12 = 6/3x

Then, we have to divide each side by 6/3 to make the x alone.

12/6/3 = 6/3x/6/3

To divide with fractions, multiply the reciprocal (opposite number on number line):

12/1·3/6 = x

36/6 = x

6 = x

Tatiana [17]3 years ago
5 0

Answer:

Step-by-step explanation:

You might be interested in
If a person tosses a coin 23 times, how many ways can he get 11 heads
EastWind [94]

Tossing a coin is a binomial experiment.

Now lets say there are 'n' repeated trials to get heads. Each of the trials can result in either a head or a tail.

All of these trials are independent since the result of one trial does not affect the result of the next trial.

Now, for 'n' repeated trials the total number of successes is given by

_{r}^{n}\textrm{C}

where 'r' denotes the number of successful results.

In our case n=23 and r=11,

Substituting the values we get,

_{11}^{23}\textrm{C}=\frac{23!}{11!\times 12!}

\frac{23!}{11!\times 12!}=1352078

Therefore, there are 1352078 ways to get heads if a person tosses a coin 23 times.


3 0
4 years ago
Equivalent to 13/5<br> A.2.4<br><br> B.2.45<br><br> C.2.55<br><br> D.2.6 <br><br> NEED HELP ASAP!!!!
jek_recluse [69]
D is the answer because 13 goes into 5 twice and 3 Is left over multiply that by two to get 6/10 than that it's equivalent to .6 so it is 2.6
7 0
3 years ago
Suppose a set of data about the spread (in hundreds of acres) of a food-borne bacteria (after t weeks) in lettuce has the follow
Alexxandr [17]

Answer:

A sinusoidal model would be used

The kind of function that have consistency in the periodic rate of change is the Average rate of changes

Step-by-step explanation:

The type of model that would be used is sinusoidal model and this is because there is periodic change in the values given ( i.e the rate of changes given )

For percentage rate of changes :

starting from 0.9% there is an increase to 1.3% then a decrease to 1.1% and a further decrease to 1% before an increase to 1.3% and another decrease to 1%

For Average rate of changes:

starting from 2.9 there is a decrease to 2.4, then an increase to 3.7 and another decrease to 3.1 followed by an increase to 3.6 and a decrease back to 3.2

This relation ( sinusoidal model ) is best suited for a linear model because there is a periodic rate of change in the functions

The kind of function that have consistency in the period rate of change is the Average rate of changes

6 0
3 years ago
31/9 as a mix number
Nataly [62]
3 and 4/9 would be the answer to this.
8 0
3 years ago
Read 2 more answers
Let X1 and X2 be independent random variables with mean μand variance σ².
My name is Ann [436]

Answer:

a) E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu

So then we conclude that \hat \theta_1 is an unbiased estimator of \mu

E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu

So then we conclude that \hat \theta_2 is an unbiased estimator of \mu

b) Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}

Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}

Step-by-step explanation:

For this case we know that we have two random variables:

X_1 , X_2 both with mean \mu = \mu and variance \sigma^2

And we define the following estimators:

\hat \theta_1 = \frac{X_1 + X_2}{2}

\hat \theta_2 = \frac{X_1 + 3X_2}{4}

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

E(\hat \theta_i) = \mu , i = 1,2

So let's find the expected values for each estimator:

E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})

Using properties of expected value we have this:

E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu

So then we conclude that \hat \theta_1 is an unbiased estimator of \mu

For the second estimator we have:

E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})

Using properties of expected value we have this:

E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu

So then we conclude that \hat \theta_2 is an unbiased estimator of \mu

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

Var(aX) = a^2 Var(X)

For the first estimator we have:

Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})

Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)]

Since both random variables are independent we know that Cov(X_1, X_2 ) = 0 so then we have:

Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}

For the second estimator we have:

Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})

Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)]

Since both random variables are independent we know that Cov(X_1, X_2 ) = 0 so then we have:

Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}

7 0
3 years ago
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