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enyata [817]
2 years ago
8

The cars access code has four digits first digit cannot be zero and the last digit must be even

Mathematics
1 answer:
arlik [135]2 years ago
4 0
There's 9 choices for the first digit and 4 choices for the last digit. The number of choices for the 2nd and 3rd digits is 90 if 2 numbers are different. or 100 if duplicates are allowed.

If duplicates are allowed the answer is 9 * 100 * 4 = 3600 possible numbers.
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Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess
Zinaida [17]

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill

\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2

4 0
3 years ago
Read 2 more answers
Plz help I need to know please.....
jarptica [38.1K]

101 feet. also -1 isnt an elevation. If you were to draw a line youd put all the negative numbers on the bottom, and positive on the top.

7 0
3 years ago
Write 12,430,000 in expanded form
lina2011 [118]
12,000,000+400,000+30,000+
8 0
3 years ago
If you borrow $894 for four years at an interest rate of 3%, how much interest will you pay?
Studentka2010 [4]

Answer:

$1006.20

Step-by-step explanation:

This is an example of compound interest.

To work this out you would first have to convert the percentage into a decimal, you can do this by dividing 3 by 100, which gives you 0.03. You are dividing by 100 as percentages are out of 100. Then you would add 1 to 0.03, which gives you 1.03. This is because you are finding percentage increases. Then you would multiply 894 by 1.03 to the power of 4, which gives you 1006.20. This is because you are finding 3 percent and adding it on and then finding 3 percent of that, this is why we put it to the power of how much time.

1) Divide 3 by 100.

3/100=0.03

2) Add 1 to 0.03.

0.03+1=1.03

3) Multiply 894 by 1.03 to the power of 4.

894*1.03^{4}=1006.20

3 0
3 years ago
Big chickens: The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1387 grams
Nataliya [291]

Answer:

a) 0.2318

b) 0.2609

c) No it is not unusual for a broiler to weigh more than 1610 grams

Step-by-step explanation:

We solve using z score formula

z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean 1387 grams and standard deviation 192 grams. Use the TI-84 Plus calculator to answer the following.

(a) What proportion of broilers weigh between 1150 and 1308 grams?

For 1150 grams

z = 1150 - 1387/192

= -1.23438

Probability value from Z-Table:

P(x = 1150) = 0.10853

For 1308 grams

z = 1308 - 1387/192

= -0.41146

Probability value from Z-Table:

P(x = 1308) = 0.34037

Proportion of broilers weigh between 1150 and 1308 grams is:

P(x = 1308) - P(x = 1150)

0.34037 - 0.10853

= 0.23184

≈ 0.2318

(b) What is the probability that a randomly selected broiler weighs more than 1510 grams?

1510 - 1387/192

= 0.64063

Probabilty value from Z-Table:

P(x<1510) = 0.73912

P(x>1510) = 1 - P(x<1510) = 0.26088

≈ 0.2609

(c) Is it unusual for a broiler to weigh more than 1610 grams?

1610- 1387/192

= 1.16146

Probability value from Z-Table:

P(x<1610) = 0.87727

P(x>1610) = 1 - P(x<1610) = 0.12273

≈ 0.1227

No it is not unusual for a broiler to weigh more than 1610 grams

8 0
3 years ago
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