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vovikov84 [41]
3 years ago
12

which one of the following examples represents a repeating decimal A. 1.111114 B. 0.777777 C. 4.252525 D. 0.123123

Mathematics
1 answer:
elena-s [515]3 years ago
7 0
Actually, i think something is missing here:

You need either a parenthesis or some dots at the end to determine this. A repeating decimal can have one repreating digit:

0.(7): 0.777777...

two:

0.(45): 0.45454545454545....

or more: so potentially all of them can be repeating, even a!

it could be: 1.(111114)

or: 1.111114111114111114111114111114111114111114111114111114111114111114111114111114...

proably B. is the most typical of repeating decimals (choosed this one if you have to), but in reality, you need more information... did you copy the question exactly?
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Nana76 [90]

Answer:

Step-by-step explanation:

Urn has 3 green and 4 yellow

we choose 2 balls randomly

so samples  space for this selection is

\left [ \left ( Y,G\right ),\left ( G,Y\right ),\left ( G,G\right ),\left ( Y,Y\right )\right ]

Event A is choosing differently colored ball form sample space

P(A)=Probability of event (G,Y)+Probability of event (Y,G)

=\frac{^3C_1\times ^4C_1}{7\times 6}+\frac{^4C_1\times ^3C_1}{7\times 6}

=\frac{4}{7}

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3 years ago
In the eighth grade,322 students voted for the new mascot to be a tiger.this was 7/10 of the total number of students in the eig
kolbaska11 [484]
322 is 7/10 of all the students.

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46 is 1/10 of all the students.

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8 0
3 years ago
Read 2 more answers
Write an equation for the following scenario.
Lelu [443]

Answer:

3.472x = 4

Step-by-step explanation:

srry if im wrong, but I tried and this question is hard

3 0
3 years ago
Blood pressure: High blood pressure has been identified as a risk factor for heart attacks and strokes. The proportion of U.S. a
Y_Kistochka [10]

Answer:

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

For this case we know this:

n=37 ,  p=0.2

We can find the standard error like this:

SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658

So then our random variable can be described as:

p \sim N(0.2, 0.0658)

Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:

P(p>0.4)

We can use the z score given by:

z = \frac{p -\mu_p}{SE_p}

And using this we got this:

P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z

And we can find this probability using the Ti 84 on this way:

2nd> VARS> DISTR > normalcdf

And the code that we need to use for this case would be:

1-normalcdf(-1000, 3.04; 0;1)

Or equivalently we can use:

1-normalcdf(-1000, 0.4; 0.2;0.0658)

Step-by-step explanation:

We need to check if we can use the normal approximation:

np = 37 *0.2 = 7.4 \geq 5

n(1-p) = 37*0.8 = 29.6\geq 5

We assume independence on each event and a random sampling method so we can conclude that we can use the normal approximation and then ,the population proportion have the following distribution :

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

For this case we know this:

n=37 ,  p=0.2

We can find the standard error like this:

SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658

So then our random variable can be described as:

p \sim N(0.2, 0.0658)

Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:

P(p>0.4)

We can use the z score given by:

z = \frac{p -\mu_p}{SE_p}

And using this we got this:

P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z

And we can find this probability using the Ti 84 on this way:

2nd> VARS> DISTR > normalcdf

And the code that we need to use for this case would be:

1-normalcdf(-1000, 3.04; 0;1)

Or equivalently we can use:

1-normalcdf(-1000, 0.4; 0.2;0.0658)

7 0
3 years ago
Using the graph below determine the slopes of the lines in simplest form.
alexandr1967 [171]

Answer:AB= -4/3

CD=6/2

EF=Zero

GH=4/6

Step-by-step explanation:

ok  rise/run

7 0
3 years ago
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