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gtnhenbr [62]
3 years ago
7

WILL GIVE BRAINLIEST anitas and caseys bills do not vary from month to month. anita pays 80 more than casey does each month. ove

r the course of 4 months,their combined bill is 1,520. write the system of equations that describes this situation and solve it to find the montly payments for anita and casey.
Mathematics
1 answer:
Mrac [35]3 years ago
8 0

Answer:

1520/4=380 per month

Step-by-step explanation:

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"a "and "b^2-4ac" must be negetive
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Use Pythagorean theorem
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Answer:

15

Step-by-step explanation:

a^2+b^2=c^2

y^2+8^2=17^2

Step 1: Simplify both sides of the equation.

y^2+64=289

Step 2: Subtract 64 from both sides.

y^2+64−64=289−64

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y=15  

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What is this answer?
Rainbow [258]

Answer:

None of the above

Step-by-step explanation:

3^3=81

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6^3=216

-6^3=-216

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3 years ago
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Carys calculates the total amount EEE, in dollars, that she earns for working hhh hours using the equation E=10hE=10hE, equals,
timurjin [86]

Answer:

  • $10 per hour
  • 0.1 hours

Step-by-step explanation:

To find Carys' earnings in one hour, fill in h=1 in the formula:

  E = 10·1 = 10

Carys earns 10 dollars per hour.

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To find the number of hours required to earn 1 dollar, fill in E=1 in the formula:

  1 = 10h

  0.1 = h . . . . . divide by 10

It takes 0.1 hours for Carys to earn 1 dollar.

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3 years ago
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A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature
Westkost [7]

Answer:

(a) The value of <em>k</em> is \frac{1}{15}.

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d)  P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5 is not the pmf of <em>y</em>.

Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of <em>k</em>.

\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}

Thus, the value of <em>k</em> is \frac{1}{15}.

(b)

Compute the value of P (Y ≤ 3) as follows:

P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\   =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5 to be the pmf of Y it has to satisfy the conditions:

  1. P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\
  2. \sum P(y)=1

<u>Check condition 1:</u>

y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0

Condition 1 is fulfilled.

<u>Check condition 2:</u>

\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1

Condition 2 is not satisfied.

Thus, P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5 is not the pmf of <em>y</em>.

7 0
3 years ago
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