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3 years ago

70% of a number is 98.

1 answer:

8 0

Set up the proportion:
70% - 98
90% - x

$\frac{70\%}{90\%}=\frac{98}{x} \\
\frac{7}{9}=\frac{98}{x} \\
\frac{7}{9}x=98 \\
x=98 \times \frac{9}{7} \\
x=14 \times 9 \\
x=126$

The answer is D. 126.

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Charra [1.4K]

9514 1404 393

Answer:

-2

Step-by-step explanation:

-8^(1/3) = -(2^3)^(1/3) = -(2^(3/3)) = -2

6 0

3 years ago

Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu

olchik [2.2K]

Answer:

a) $s^2 =30^2 =900$

b) $\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

c) $23.818 \leq \sigma \leq 41.112$

Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

$s^2 =30^2 =900$

Part b

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom are given by:

$df=n-1=20-1=19$

Since the Confidence is 0.90 or 90%, the significance $\alpha=0.1$ and $\alpha/2 =0.05$ , the critical values for this case are:

$\chi^2_{\alpha/2}=30.144$

$\chi^2_{1- \alpha/2}=10.117$

And replacing into the formula for the interval we got:

$\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

Part c

Now we just take square root on both sides of the interval and we got:

$23.818 \leq \sigma \leq 41.112$

5 0

4 years ago

Solve the following system of equations for a and for b:

viktelen [127]

Answer:

$a=3\\b=1$

Step-by-step explanation:

$9a+3b=30\\8a+4b=28$

Let's solve the second equation for a to later on replace it in the first equation.

$8a+4b=28\\8a=28-4b\\a=\frac{28-4b}{8}$

Now plug this into the first equation.

$9a+3b=30\\9(\frac{28-4b}{8})+3b=30$

Distribute the 9

$(\frac{252-36b}{8}) +3b=30$

Break down the fraction.

$\frac{252}{8}-\frac{36b}{8}+3b=30$

Simplify.

$\frac{63}{2}-\frac{9}{2}b+3b=30$

Subtract $\frac{63}{2}$

$-\frac{9}{2}b+3b=30-\frac{63}{2}$

Combine like terms.

$\frac{-9+2*3}{2}b=\frac{30*2-63}{2}$

$\frac{-9+6}{2}b=\frac{60-63}{2}$

$\frac{-3}{2}b=\frac{-3}{2}$

Muliply by the reciprocal or inverted fraction next to b.

$(-\frac{2}{3})(-\frac{3}{2}) b=-\frac{3}{2}(-\frac{2}{3})$

$b=1$

Now plug this value into any of the equations to find the value of a.

$8a+4b=28\\8a+4(1)=28\\8a+4=28\\8a=28-4\\8a=24\\a=\frac{24}{8}\\ a=3$

5 0

4 years ago

If u see this forget it, I typed it in the wrong spot

Arte-miy333 [17]

Ok then um so ya ok thx for the points though

4 0

3 years ago

Someone please help me make a proof for the bottom question

Lerok [7]

Answer:

DB = CA (Proved)

Step-by-step explanation:

Statement 1.

∠D = ∠C, M is the midpoint of DC and ∠1 = ∠2

Reason 1.

Given

Statement 2.

Between Δ DBM and Δ CAM,

(i) DM = CM,

(ii) ∠D = ∠C and

(iii) ∠DMB = ∠CMA

Reason 2.

(i) given

(ii) given and

(iii) ∠ DMB = ∠1 + ∠AMB and ∠CMA = ∠2 + ∠AMB

Since ∠1 = ∠2, so, ∠DMB = ∠CMA.

Statement 3.

Δ DBM ≅ Δ CAM

Reason 3.

By angle-side-angle rule.

Statement 4.

DB = CA

Reason 4.

Corresponding sides of two congruent triangles. (Answer)

7 0

3 years ago