Please help image attached

A music store charges $18.50 for a CD plus a 0.5% city tax and a separate 4% state tax.

Strike441 [17]

A) The equation will become: 18.50 (1.005) + 18.50 (0.04)= 19.3325
Which can be rounded to 19.33
b) 18.50 (1.04)-18.50(1.005)= 0.6475
Which can be rounded to 0.65

Therefore your answer is Number 1)

3 0

3 years ago

5. Molly has 5 sticks of lengths 2,4,6,8, and 10 inches. Using three sticks at 1 point

sergeinik [125]

Answer:

4, 6 & 8 / 4, 8 & 10 , / 6, 8 & 10

So 3 triangles

Step-by-step explanation:

The longest side of a triangle has to be greater than the difference of the other 2 sides, and less than the sum of the other 2.

diff < long side < sum

Not equal to, less than & greater than.

6 0

2 years ago

The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)

Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ .

Step-by-step explanation:

We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

Let A = Math Scores ,B = Writing Scores and D = difference between both

So, $\mu_A$ = Population mean for the math scores

$\mu_B$ = Population mean for the writing scores

Let $\mu_D$ = Difference between the population mean for the math scores and the population mean for the writing scores.

Null Hypothesis, $H_0$ : $\mu_A = \mu_B$ or $\mu_D$ = 0

Alternate Hypothesis, $H_1$ : $\mu_A \neq \mu_B$ or $\mu_D \neq$ 0

Hence, Test Statistics used here will be;

$\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ where, Dbar = Bbar - Abar

$s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$

n = 12

Student Math scores (A) Writing scores (B) D = B - A

1 540 474 -66

2 432 380 -52

3 528 463 -65

4 574 612 38

5 448 420 -28

6 502 526 24

7 480 430 -50

8 499 459 -40

9 610 615 5

10 572 541 -31

11 390 335 -55

12 593 613 20

Now Dbar = Bbar - Abar = 489 - 514 = -25

Bbar = $\frac{\sum B_i}{n}$ = $\frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}$ = 489

Abar = $\frac{\sum A_i}{n}$ = $\frac{540+432+528+574+448+502+480+499+610+572+390+593}{12}$ = 514

∑ $D_i^{2}$ = 22600 and $s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$ = $\sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} }$ = 37.05

So, Test statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$

= $\frac{-25 - 0}{\frac{37.05}{\sqrt{12} } }$ follows $t_1_1$ = -2.34

Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

8 0

3 years ago

Solve for I when 3/5 +I-1/15

Ugo [173]

Answer:

23 over 15 or 1.533

Step-by-step explanation:

Group like terms:

1+(3/5+-1/15)

Find LCD:

1+9/15+-1/15

Combine numerators:

15/15+8/15

And you get 23/15

Simplified:1 8/15

Decimal: 1.533

4 0

2 years ago

A square ceiling has an area of 360 square feet. Alex wants to put molding around the entire perimeter of the ceiling. Approxima

Blababa [14]

Answer:

Option C is correct

76 feet

Step-by-step explanation:

Area of square(A) is given by:

$A =s^2$

where, s is the side of the square.

As per the statement:

A square ceiling has an area of 360 square feet.

⇒A = 360 square feet.

then;

$s^2 = 360$

⇒ $s = \sqrt{360} =18.973666$

Perimeter(P) of square is given by:

$P = 4s$

then;

$P = 4 \cdot 18.973666 = 75.894664 \approx 76 ft$

Therefore, 76 feet molding does he need.

7 0

3 years ago