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Neko [114]
2 years ago
10

Barry’s 6-digit passcode for internet banking consists of six different digits. If a multiplication sign is placed between the s

econd and third digits and an equals sign is placed between the third and fourth digits, a correct calculation appears.
Mathematics
1 answer:
elena-14-01-66 [18.8K]2 years ago
4 0

Answer:

This question i think google could help you if not more then brainy if u dont get a answer fast enough

Step-by-step explanation:

hope i helped u a little bit i know i didn't have a answer.

You might be interested in
Group of baseball fans can see home plate from a 40 meter tall building outside the stadium. The angle of vision has a tangent o
dimulka [17.4K]

Answer:

17.78 meters

Step-by-step explanation:

Let

x ----> the horizontal distance, in meters, to home plate

\theta ----> the angle of vision

we know that

tan(\theta)=\frac{40}{x} ----> by TOA (opposite side divided by the adjacent side)

we have

tan(\theta)=\frac{9}{4}

substitute

\frac{9}{4}=\frac{40}{x}

solve for x

x=40(4)/9\\x=17.78\ m

5 0
2 years ago
A certain article indicates that in a sample of 1,000 dog owners, 610 said that they take more pictures of their dog than of the
lbvjy [14]

Answer:

(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).

(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).

(c) First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

Step-by-step explanation:

(a)

Let <em>X</em> = number of dog owners who take more pictures of their dog than of their significant others or friends.

Given:

<em>X</em> = 610

<em>n</em> = 1000

Confidence level = 90%

The (1 - <em>α</em>)% confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

The sample proportion is:

\hat p=\frac{X}{n}=\frac{610}{1000}=0.61

The critical value of <em>z</em> for a 90% confidence level is:

z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645

Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)

Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).

<u>Interpretation</u>:

There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).

Correct option is (3).

(b)

Let <em>X</em> = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

<em>X</em> = 440

<em>n</em> = 1000

Confidence level = 95%

The (1 - <em>α</em>)% confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

The sample proportion is:

\hat p=\frac{X}{n}=\frac{440}{1000}=0.44

The critical value of <em>z</em> for a 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)

Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend  is (0.42, 0.46).

<u>Interpretation</u>:

There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).

Correct option is (1).

(c)

The confidence interval in part (b) is wider than the confidence interval in part (a).

The width of the interval is affected by:

  1. The confidence level
  2. Sample size
  3. Standard deviation.

The confidence level in part (b) is more than that in part (a).

Because of this the critical value of <em>z</em> in part (b) is more than that in part (a).

Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.

First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

4 0
1 year ago
The graphs below have the same shape. What is the equation of the blue
Soloha48 [4]
It would be (x-4)^2

It has a horizontal shift of 4 to the right
6 0
1 year ago
Read 2 more answers
Suppose the time required for an auto shop to do a tune-up is normally distributed, with a mean of 102 minutes and a standard de
hammer [34]

Answer:

Step-by-step explanation:

Suppose the time required for an auto shop to do a tune-up is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = points scored by students

u = mean time

s = standard deviation

From the information given,

u = 102 minutes

s = 18 minutes

1) We want to find the probability that a tune-up will take more than 2hrs. It is expressed as

P(x > 120 minutes) = 1 - P(x ≤ 120)

For x = 120

z = (120 - 102)/18 = 1

Looking at the normal distribution table, the probability corresponding to the z score is 0.8413

P(x > 120) = 1 - 0.8413 = 0.1587

2) We want to find the probability that a tune-up will take lesser than 66 minutes. It is expressed as

P(x < 66 minutes)

For x = 66

z = (66 - 102)/18 = - 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.02275

P(x < 66 minutes) = 0.02275

4 0
2 years ago
For a package to qualify for a certain postage rate, the sum of its length and girth cannot exceed 85 inches. If the girth is 63
tester [92]

Answer:

Length of package can be = 22 inches

Step-by-step explanation:

Given:

For a package to qualify for a certain postage rate, the sum of its length and girth cannot exceed 85 inches

To find length of package when girth of package is = 63 inches

Solution:

Let length of package be = l inches

Let girth length of package be = g inches

Sum of length and girth of package = (l+g) inches  

To qualify for a certain postage rate the sum of length and girth should not exceed 85 inches.

Thus, the inequality representing the situation can be given as:

l+g\leq 85

We are given girth of package is = 63 inches

So, inequality to find length would be:

l+63\leq 85

Subtracting both sides by 63.

l+63-63\leq 85-63

l\leq 22 inches

So, length should not exceed 22 inches in order to qualify.

Thus, the maximum length of package to qualify for the postal rate must be = 22 inches

3 0
2 years ago
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