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If 30 Coins In Quarters And Dimes Are Worth $4.80, How Many Of Each Coin Are There?

GREYUIT [131]

Answer:

There are 18 dimes and 12 quarters

Step-by-step explanation:

Remember that

1 Dime =$0.10

1 Quarter=$0.25

Let

x-----> the number of dimes

y-----> the number of quarters

we know that

x+y=30

x=30-y ------> equation A

0.10x+0.25y=4.80

Multiply by 100 both sides

10x+25y=480 ------> equation B

Substitute equation A in equation B and solve for y

10(30-y)+25y=480

300-10y+25y=480

15y=180

y=12 quarters

Find the value of x

x=30-y=30-12=18 dimes

therefore

There are 18 dimes and 12 quarters

3 0

3 years ago

whitch integer is described by the following? The absolute value is 3 and the number is to the left of 0 on the number line

levacccp [35]

-3 because absolute value is basically any number between two lines like this l -3 l the absolute value of -3 is three.

4 0

3 years ago

What are the coordinates of the point on the directed line segment from

ziro4ka [17]

Answer:

The desired point is thus (-1/2, -5).

Step-by-step explanation:

The x-component of this directed line segment is 2 - (-8), or 10, and the y-component is -7 - (1), or -8. This segment is in Quadrant II, since the x-component is positive and the y-component is negative.

The point of interest is (3/4) of the way in the positive x-direction from x = -8. We can express this symbolically as -8 + (3/4)(10), or -8 + 7.5, or -1/2.

The point of interest is 3/4 of the way in the negative y direction from 1, or:

1 + (3/4)(-8), or 1 - 6, or -5.

The desired point is thus (-1/2, -5).

3 0

2 years ago

if 42% of a crowd of 3850 at a football match were females, how many females attended the football game?

sweet [91]

Answer: 1617

Step-by-step explanation:

You would need to find 42% of 3850

Convert 42% into decimal:

42/100 = 0.42

Multiply:

0.42*3850

= 1617

Therefore, there were 1617 females that attended the game

7 0

3 years ago

Read 2 more answers

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