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musickatia [10]
3 years ago
7

Kendra has $ 7.80 $7.80 in dimes and quarters. if she has 2 2 times as many quarters as dimes, how many of each coin does she ha

ve?
Mathematics
1 answer:
alexgriva [62]3 years ago
7 0
Kendra would have 26 quarters and 13 dimes.
26x.25=6.5
26/2=13
13x.1=1.3
6.5+1.3=7.8
You might be interested in
[(7+3)x5-4](divided by) 2+3<br> SHOW ALL WORK WILL MARK BRAINLIEST
Nesterboy [21]

Answer:

26

Step-by-step explanation:

[(7+3)5-4]/2+3

-To solve this equation you have to use PEMDAS

P- Parentheses

E- Exponents

M- Multiplication

D- Division

A- Addition

S- Subtraction-

- With MD and AS you work left to right of the equation since they are in the same spot. (PE[MD][AS])

Step 1) [(10)5-4]/2+3

- First you do "P," parentheses, so you add 7+3=10

Step 2) [50-4]/2+3

- Next you do "M," multiplication, and multiply 10x5=50

Step 3) [46]/2+3

- Then you do "S," subtraction, and subtract 50-4=46

(FYI: Steps 1-3 were still in the parentheses. We had to start with the parentheses in the parentheses, work PEMDAS, and now we are out of the parentheses and have to work PEMDAS on the rest of the problem.)

Step 4) 23+3

- Now we do "D," division, and divide 46/2=23

Step 5) 23+3=6

- Finally we do "A," addition, and add 23+3=26 so the answer is 26

(FYI: "/" means division)

4 0
2 years ago
Help for brianilist please!!!
Dima020 [189]
Hi there!!!

So using this rule:

Area = base * height
20ft  * 15ft  = 300ft^2
↑            ↑
base    height      

HOPE IT HELPS U!!!!!!!!!
THE ANSWER IS 300 FT^2

~ TRUE BOSS




5 0
3 years ago
The heights of men in a certain population follow a normal distribution with mean 69.7 inches and standard deviation 2.8 inches.
Mama L [17]

Answer:

a) P(Y > 76) = 0.0122

b) i) P(both of them will be more than 76 inches tall) = 0.00015

   ii) P(Y > 76) = 0.0007

Step-by-step explanation:

Given - The heights of men in a certain population follow a normal distribution with mean 69.7 inches and standard deviation 2.8 inches.

To find - (a) If a man is chosen at random from the population, find

                    the probability that he will be more than 76 inches tall.

              (b) If two men are chosen at random from the population, find

                    the probability that

                    (i) both of them will be more than 76 inches tall;

                    (ii) their mean height will be more than 76 inches.

Proof -

a)

P(Y > 76) = P(Y - mean > 76 - mean)

                 = P( \frac{( Y- mean)}{S.D}) > \frac{( 76- mean)}{S.D})

                 = P(Z >  \frac{( 76- mean)}{S.D})

                 = P(Z > \frac{76 - 69.7}{2.8})

                 = P(Z > 2.25)

                 = 1 - P(Z  ≤ 2.25)

                 = 0.0122

⇒P(Y > 76) = 0.0122

b)

(i)

P(both of them will be more than 76 inches tall) = (0.0122)²

                                                                           = 0.00015

⇒P(both of them will be more than 76 inches tall) = 0.00015

(ii)

Given that,

Mean = 69.7,

\frac{S.D}{\sqrt{N} } = 1.979899,

Now,

P(Y > 76) = P(Y - mean > 76 - mean)

                 = P( \frac{( Y- mean)}{\frac{S.D}{\sqrt{N} } })) > \frac{( 76- mean)}{\frac{S.D}{\sqrt{N} } })

                 = P(Z > \frac{( 76- mean)}{\frac{S.D}{\sqrt{N} } })

                 = P(Z > \frac{( 76- 69.7)}{1.979899 }))

                 = P(Z > 3.182)

                 = 1 - P(Z ≤ 3.182)

                 = 0.0007

⇒P(Y > 76) = 0.0007

6 0
2 years ago
Identify the type of conic section that has the equation 9x2+ 16y2 = 144 and identify its domain and range.
Kobotan [32]
9 x² + 16 y² = 144  /:144
\frac{ x^{2} }{16} + \frac{ y^{2} }{9} =1
General formula of ellipse ( the center is at the origin ):\frac{ x^{2} }{ a^{2} }+ \frac{ y^{2} }{ b^{2} } =1 
a² = 16,  b² = 9
Domain: [-a, a ] = [-4, 4]
Range:[-b, b ]
Answer: B ) ellipse.
Domain: { -4 ≤ x ≤ 4 }
Range: { -3 ≤ y ≤ 3 }

 
8 0
2 years ago
The graphs below have the same shape. What is the equation of the red
anyanavicka [17]

Answer:

Option B is correct .

Step-by-step explanation:

According to Question , both the graph have same shape . If we look at the the first graph it cuts x - axis at (0 , 2) and ( 0 , -2) . Hence x = 2 and -2 are the zeroes of the equation .

And ,the given function is ,

\implies f(x) = 4 - x^2 \\\\\implies f(x) = 4-x^2=0 \\\\\implies 2^2-x^2=0\\\\\implies (2-x)(2+x) = 0 \\\\\boxed{\red{\bf \implies x = 2 , (-2) }}

<u>Hence ,we can can see that x = </u><u> </u><u>2</u><u> </u><u>and</u><u> </u><u>(</u><u>-</u><u>2</u><u>)</u><u> </u><u>are</u><u> </u><u>the</u><u> </u><u>zeroes </u><u>of </u><u>graph</u><u>. </u><u> </u>

This implies that if we know the zeroes , we can frame the Equation.

On looking at second parabola , it's clear that cuts x - axis at ( 1, 0 ) and (-1,0). So , 1 and -1 are the zeroes of the quadratic equation . Let the function be g(x) . Here , a and ß are the zeroes.

\implies g(x) = (x-\alpha)(x-\beta) \\\\\implies g(x) = (1+x)(1-x) \\\\\boxed{\pink{\bf \implies g(x) = 1 - x^2}}

<u>Hence </u><u>option </u><u>B</u><u> </u><u>is</u><u> </u><u>corre</u><u>ct</u><u> </u><u>.</u>

8 0
2 years ago
Read 2 more answers
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