Solve this equation q - 17 = 18<br><br> Thanks!

A theater group made appearances in two cities. The hotel charges before tax in the second city were $1500 lower than in the fir

KonstantinChe [14]

Answer:

The hotel bill in the first city before tax would be $ 650

Step-by-step explanation:

Let x be the hotel charges ( in dollars ) before tax in the first city,

∵ Hotel charges before tax in the second city were $1500 lower than in the first.

So, the hotel charges in second city = ( x - 1500 ) dollars,

Now, percentage of tax in,

First city = 8%

Second city = 10%,

So, the total tax in both city = 8% of x + 10% of (x-1500)

$=\frac{8x}{100}+\frac{10(x-1500)}{100}$

$=0.8x + 0.10(x-1500)$

According to the question,

$0.8x + 0.10(x-1500)=435$

$0.8x + 0.10x - 150 = 435$

$0.9x = 435 + 150$

$0.9x = 585$

$\implies x = \frac{585}{0.9}=650$

Hence, the hotel bill in the first city before tax would be $ 650.

7 0

3 years ago

Read 2 more answers

Over the past several years, the proportion of one-person households has been increasing. The Census Bureau would like to test t

bixtya [17]

Answer:

For this case we can find the critical value with the significance level $\alpha=0.05$ and if we find in the right tail of the z distribution we got:

$z_{\alpha}= 1.64$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing we got:

$z=\frac{0.344 -0.27}{\sqrt{\frac{0.27(1-0.27)}{125}}}=1.86$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of households with one person is significantly higher than 0.27

Step-by-step explanation:

We have the following dataset given:

$X= 43$ represent the households consisted of one person

$n= 125$ represent the sample size

$\hat p= \frac{43}{125}= 0.344$ estimated proportion of households consisted of one person

We want to test the following hypothesis:

Null hypothesis: $p \leq 0.27$

Alternative hypothesis: $p>0.27$

And for this case we can find the critical value with the significance level $\alpha=0.05$ and if we find in the right tail of the z distribution we got:

$z_{\alpha}= 1.64$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing we got:

$z=\frac{0.344 -0.27}{\sqrt{\frac{0.27(1-0.27)}{125}}}=1.86$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of households with one person is significantly higher than 0.27

5 0

3 years ago

Help me solve this problem please

Sophie [7]

Answer:

8x - 10

Step-by-step explanation:

3x - 7 + x + 2 + 3x - 7 + x + 2

8x - 10

8 0

4 years ago

A 3-mi cab ride costs $7.30. A 8-mi cab ride costs $15.30. Find a linear equation that models a relationship between cost c and

Rus_ich [418]

Answer:

Option c) $c=1.60d+2.5$

Step-by-step explanation:

Let

d -----> the distance in miles

c ----> the cost in dollars

we have the ordered pairs

(3,7.30) and (8,15.30)

step 1

Find the slope of the linear equation

The slope is equal to

$m=\frac{15.30-7.30}{8-3}$

$m=\frac{8}{5}$

step 2

Find the equation of the line in point slope form

$c-c1=m(d-d1)$

we have

$point\ (3,7.30)$

$m=\frac{8}{5}$

substitute

$c-7.30=\frac{8}{5}(d-3)$

$c=\frac{8}{5}d-\frac{24}{5}+7.30$

$c=1.60d+2.5$

4 0

4 years ago

A company did a quality check on all the packs of nuts it manufactured. Each pack of nuts is targeted to weigh 18.25 oz. A pack

IrinaVladis [17]

Answer:

$x < 17.89$ or $x > 18.61$ , since $\left | x-18.25 \right |>0.36$

Step-by-step explanation:

According to the question the weight is rejected if it is more than 0.36 oz, the weight should not be beyond 0.36 oz that is from the target weight of 18.25.

We can say that, the difference between the actual weight and the target weight is greater than 0.36 oz.

Lets say the actual weight is 'x'. So,

$x-18.25>0.36$