Solve for x<br><br> 1/3x=7/6 + 1/9

Plz help me with my algebra

Rus_ich [418]

The answer is a(x)=x+3

3 0

3 years ago

Read 2 more answers

Х- 1, 2, 3 ,4

Len [333]

I think it’s C, good luck

4 0

3 years ago

Addy bought a new shirt and she had a 20% off coupon . The coupon took $8.25 off of the price of her shirt. What did she pay wit

zloy xaker [14]

Answer:

$33

Step-by-step explanation:

Given: Addy had 20% off coupon.

The coupon took $8.25 off of the shirt price.

Lets assume the original cost of shirt be "x".

We know Addy had 20% off coupon, which help to reduce the price of shirt by $8.25.

∴ We can form an equation to know the original price of shirt.

⇒ $\frac{20}{100} \times x= \$ 8.25$

⇒ $\frac{1}{5} \times x= \$ 8.25$

Multiplying both side by 5

⇒ $x= 8.25\times 5$

∴ $x= \$ 41.25$

Hence, The original cost price of shirt was $41.25.

Next, finding the price paid by Addy for her new shirt by reducing coupon discount from the cost of shirt.

Price paid by Addy for her new shirt= $\$ 41.25- \$ 8.25= \$ 33$

Hence, Addy paid $33 for her new shirt.

8 0

3 years ago

My brother wants to estimate the proportion of Canadians who own their house.What sample size should be obtained if he wants the

AVprozaik [17]

Answer:

a) $n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07$

And rounded up we have that n=1476

b) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$

And rounded up we have that n=1681

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by: