4) 3 + 2x - 8x = -15

What is the distance formula?

igomit [66]

Answer:

Distance Formula is

Step-by-step explanation:

d=√((x_2-x_1)²+(y_2-y_1)²)

3 0

2 years ago

Read 2 more answers

I toss an unfair coin 12 times. This coin is 65% likely to show up heads. Calculate the probability of the following.

mars1129 [50]

Answer:

a. 0.0368

b. 0.99992131

c. 0.2039

d. 0.0048

e. 0.6533

Step-by-step explanation:

Let the probability of obtaining a head be p = 65% = 13/20 = 0.65. The probability of not obtaining a head is q = 1 - p = 1 -13/20 = 7/20 = 0.35

Since this is a binomial probability, we use a binomial probability.

a. The probability of obtaining 11 heads is ¹²C₁₁p¹¹q¹ = 12 × (0.65)¹¹(0.35) = 0.0368

b. Probability of 2 or more heads P(x ≥ 2) is

P(x ≥ 2) = 1 - P(x ≤ 1)

Now P(x ≤ 1) = P(0) + P(1)

= ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹

= (0.65)⁰(0.35)¹² + 12(0.65)¹(0.35)¹¹

= 0.000003379 + 0.00007531

= 0.0007869

P(x ≥ 2) = 1 - P(x ≤ 1)

= 1 - 0.00007869

= 0.99992131

c. The probability of obtaining 7 heads is ¹²C₇p⁷q⁵ = 792(0.65)⁷(0.35)⁵ = 0.2039

d. The probability of obtaining 7 heads is ¹²C₉q⁹p³ = 220(0.65)³(0.35)⁹ = 0.0048

e. Probability of 8 heads or less P(x ≤ 8) = ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹ + ¹²C₂p²q¹⁰ + ¹²C₃p³q⁹ + ¹²C₄p⁴q⁸ + ¹²C₅p⁵q⁷ + ¹²C₆p⁶q⁶ + ¹²C₇p⁷q⁵ + ¹²C₈p⁸q⁴

= = ¹²C₀(0.65)⁰(0.35)¹² + ¹²C₁(0.65)¹(0.35)¹¹ + ¹²C₂(0.65)²(0.35)¹⁰ + ¹²C₃(0.65)³(0.35)⁹ + ¹²C₄(0.65)⁴(0.35)⁸ + ¹²C₅(0.65)⁵(0.35)⁷ + ¹²C₆(0.65)⁶(0.35)⁶ + ¹²C₇(0.65)⁷(0.35)⁵ + ¹²C₈(0.65)⁸(0.35)⁴

= 0.000003379 + 0.00007531 + 0.0007692 + 0.004762 + 0.01990 + 0.05912 + 0.1281 + 0.2039 + 0.2367

= 0.6533

3 0

3 years ago

A large bag contains 5/12 pounds of nails. How many 1/8 pounds bags can be filled with this amount of nails?

Amanda [17]

so, three bags can be filled and 0.33 or 0.33*(1/3) =11/100 nails will be left

Step-by-step explanation:

We will get the number of bags that can be filled from 5/12 by dividing the total amount by the 1/8

Given

Total nails in bag = 5/12 pounds

The capacity of small bag = 1/8 pounds

So,

$Number\ of\ bags = \frac{\frac{5}{12}}{\frac{1}{8}}\\Simplifying\\=\frac{5}{12}*\frac{8}{1}\\=\frac{40}{12}\\=3.33\ bags$

so, three bags can be filled and 0.33*(1/3) = 11/100 nails will be left

Keywords: Division, Quotient

Learn more about conversions at:

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4 0

2 years ago

Please give me answers

Nikolay [14]

For number 9. its 34 × 65 = 65 × 34.. So that means its communicative property

6 0

3 years ago

Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver’s license. Twenty-five years later (in 2008) that perc

Dima020 [189]

Answer:

a) $ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019$

b) $ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245$

c) On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of $\hat p$ the margin of error change for part a and b.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

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

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=\pm 1.96$

If we replace the values into equation (a) for 1983 we got:

$ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019$

Part b

Since is the same confidence level the z value it's the same.

If we replace the values into equation (a) for 2008 we got:

$ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245$

Is the margin of error the same in parts (a) and (b)? Why or why not?

On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of $\hat p$ the margin of error change for part a and b.

3 0

3 years ago