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3 years ago

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Brody is paying back his father for an interest-free car loan. Brody owes his father $6000 dollars. The chart shows the cumulati

ve total he has paid his father at the end of each month. After what month will his loan be paid in full? A) December B) February C) January D) March

2 answers:

natita [175]3 years ago

5 0

March

Step-by-step explanation:

erik [133]3 years ago

4 0

How much is he paying a month?

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Help plz i will make u a brainllest 24.Can you use the SSS Postulate or the SAS Postulate to prove abd=dca? 26 28

Alex_Xolod [135]

Answer: 24) C. SSS only

26) D. DH = HF

<u>Step-by-step explanation:</u>

24.

<u>Statement</u> <u>Reason</u>

1. BD = CA 1. Given

2. AB = CD 2. Given

3. AD = AD 3. Reflexive Property

4. ΔABD = ΔDCA 4. SSS Congruency Theorem

<em>We have no information about the angles so cannot use SAS Theorem.</em>

26. G H F

E H D

Line up the letters to find the congruent sides:

⇒ GH = EH

HF = HD

GF = ED

7 0

3 years ago

Suppose there are 4 defective batteries in a drawer with 10 batteries in it. A sample of 3 is taken at random without replacemen

SSSSS [86.1K]

Answer:

a.) 0.5

b.) 0.66

c.) 0.83

Step-by-step explanation:

As given,

Total Number of Batteries in the drawer = 10

Total Number of defective Batteries in the drawer = 4

⇒Total Number of non - defective Batteries in the drawer = 10 - 4 = 6

Now,

As, a sample of 3 is taken at random without replacement.

a.)

Getting exactly one defective battery means -

1 - from defective battery

2 - from non-defective battery

So,

Getting exactly 1 defective battery = ⁴C₁ × ⁶C₂ = $\frac{4!}{1! (4 - 1 )!}$ × $\frac{6!}{2! (6 - 2 )!}$

= $\frac{4!}{(3)!}$ × $\frac{6!}{2! (4)!}$

= $\frac{4.3!}{(3)!}$ × $\frac{6.5.4!}{2! (4)!}$

= $4$ × $\frac{6.5}{2.1! }$

= $4$ × $15$ = 60

Total Number of possibility = ¹⁰C₃ = $\frac{10!}{3! (10-3)!}$

= $\frac{10!}{3! (7)!}$

= $\frac{10.9.8.7!}{3! (7)!}$

= $\frac{10.9.8}{3.2.1!}$

= 120

So, probability = $\frac{60}{120} = \frac{1}{2} = 0.5$

b.)

at most one defective battery :

⇒either the defective battery is 1 or 0

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 0 , then 3 non defective

Possibility = ⁴C₀ × ⁶C₃

= $\frac{4!}{0! (4 - 0)!}$ × $\frac{6!}{3! (6 - 3)!}$

= $\frac{4!}{(4)!}$ × $\frac{6!}{3! (3)!}$

= 1 × $\frac{6.5.4.3!}{3.2.1! (3)!}$

= 1× $\frac{6.5.4}{3.2.1! }$

= 1 × 20 = 20

getting at most 1 defective battery = 60 + 20 = 80

Probability = $\frac{80}{120} = \frac{8}{12} = 0.66$

c.)

at least one defective battery :

⇒either the defective battery is 1 or 2 or 3

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 2 , then 1 non defective

Possibility = ⁴C₂ × ⁶C₁

= $\frac{4!}{2! (4 - 2)!}$ × $\frac{6!}{1! (6 - 1)!}$

= $\frac{4!}{2! (2)!}$ × $\frac{6!}{1! (5)!}$

= $\frac{4.3.2!}{2! (2)!}$ × $\frac{6.5!}{1! (5)!}$

= $\frac{4.3}{2.1!}$ × $\frac{6}{1}$

= 6 × 6 = 36

If the defective battery is 3 , then 0 non defective

Possibility = ⁴C₃ × ⁶C₀

= $\frac{4!}{3! (4 - 3)!}$ × $\frac{6!}{0! (6 - 0)!}$

= $\frac{4!}{3! (1)!}$ × $\frac{6!}{(6)!}$

= $\frac{4.3!}{3!}$ × 1

= 4×1 = 4

getting at most 1 defective battery = 60 + 36 + 4 = 100

Probability = $\frac{100}{120} = \frac{10}{12} = 0.83$

3 0

3 years ago

AYUDAAA porfavorrrr primera urna 6 bolas verdes y 3 rojas segunda urna 3 verdes, 3 blancas y 3 rojas tercera urna 6 verdes, 1 bl

marin [14]

We have the following information:

first urn: 6 green balls and 3 red ones

total: 6 + 3 = 9

second urn: 3 green, 3 white and 3 red

total: 3 + 3 + 3 = 9

third urn: 6 green, 1 white and 2 red

total: 6 + 1 + 2 = 9

a) A green ball is more likely to be obtained, since there are more green balls than red balls, which makes the probability higher.

b) probability of drawing a green, red and white ball.

first urn:

green = 6/9 = 66.66%

red = 3/9 = 33.33%

white = 0/9 = 0%

second urn:

green = 3/9 = 33.33%

red = 3/9 = 33.33%

white = 3/9 = 33.33%

third urn:

green = 6/9 = 66.66%

red = 2/9 = 22.22%

white = 1/9 = 11.11%

c) it would be chosen where the probability of drawing green would be the highest, which means that it would be possible both in the first and in the third ballot box, the probability is equal 66.66%

d) without a green ball, the third ballot box would look like this:

5 green balls, 2 red balls and 1 white ball, with a total of 8.

The probability of drawing would be:

green = 5/8 = 62.5%

red = 2/8 = 25%

white = 1/8 = 12.5%

7 0

3 years ago

What type of graph is this?

murzikaleks [220]

Answer:

The graph of a quadratic equation, or a parabola, looks like a U, an upside down U, a C, or a backwards C. We can use the following rules to determine what the graph of a given quadratic equation looks like. If y = ax2 + bx + c, and a is positive, then the graph of the equation is the shape of a U.

Step-by-step explanation:

bc

5 0

3 years ago

Can someone help me with this math problem ?? Pls it’s for my final !

Airida [17]

Answer:

221.87 feet

Step-by-step explanation:

Given that,

A 525 ft cable runs from the top of an antenna to the ground.

The angle of elevation made by the ground to the top of an antena 25°

We need to find the height of the antenna.

Using trigonometry,

Hypotenuse, H = 525 ft

θ = 25°

So,

$\sin\theta=\dfrac{P}{H}\\\\P=H\sin\theta\\\\P=525\times \sin(25)\\\\P=221.87\ ft$

So, the height of the antenna is equal to 221.87 feet.

4 0

3 years ago