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3 years ago

5

Karen divided her books and put them on 6 shelves. There were at least 14 books on each shelf. How many books did she have?? Wri

te an inequality to represent this situation.

2 answers:

dem82 [27]3 years ago

4 0

What you do is 14 times 6 and then you have your answer

andrew11 [14]3 years ago

3 0

14*6 is less than or equal to 84

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A small-business owner will lease a billboard to advertise her business. The table shows the linear relationship between the cos

soldi70 [24.7K]

It would take about 25 months for the cost of leasing a billboard be the same for both companies.

<h3>Linear function</h3>

A linear function is in the form:

y = mx + b

Where y,x are variables, m is the rate of change and b is the initial value of y.

Let P represent the cost to lease a billboard from a different company for x months. Given that:

P(x) = 100x

For the second company, let C represent the cost, hence using the table point (1, 580) and (5, 900):

C - 580 = [(900 - 580)/(5 - 1)](x - 1)

C = 80x + 500

For the same cost:

100x = 80x + 500

x = 25 months

It would take about 25 months for the cost of leasing a billboard be the same for both companies.

Find out more on Linear function at: brainly.com/question/15602982

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2 years ago

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3 years ago

Name five equivalent fractions to 5/10

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3 years ago

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3 years ago

The following observations were made on fracture toughness of a base plate of 18% nickel maraging steel (in ksi √in, given in in

Grace [21]

Answer:

A 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].

Step-by-step explanation:

We are given the following observations that were made on fracture toughness of a base plate of 18% nickel maraging steel below;

68.6, 71.9, 72.6, 73.1, 73.3, 73.5, 75.5, 75.7, 75.8, 76.1, 76.2, 76.2, 77.0, 77.9, 78.1, 79.6, 79.8, 79.9, 80.1, 82.2, 83.7, 93.4.

Firstly, the pivotal quantity for finding the confidence interval for the standard deviation is given by;

P.Q. = $\frac{(n-1) \times s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation = $\sqrt{\frac{\sum (X - \bar X^{2}) }{n-1} }$ = 5.063

$\sigma$ = population standard deviation

n = sample of observations = 22

Here for constructing a 90% confidence interval we have used One-sample chi-square test statistics.

<u>So, 90% confidence interval for the population standard deviation, </u> $\sigma$ <u> is ;</u>

P(11.59 < $\chi^{2}__2_1$ < 32.67) = 0.90 {As the critical value of chi at 21 degrees

of freedom are 11.59 & 32.67}

P(11.59 < $\frac{(n-1) \times s^{2} }{\sigma^{2} }$ < 32.67) = 0.90

P( $\frac{ 11.59}{(n-1) \times s^{2}}$ < $\frac{1}{\sigma^{2} }$ < $\frac{ 32.67}{(n-1) \times s^{2}}$ ) = 0.90

P( $\frac{(n-1) \times s^{2} }{32.67 }$ < $\sigma^{2}$ < $\frac{(n-1) \times s^{2} }{11.59 }$ ) = 0.90

<u>90% confidence interval for</u> $\sigma^{2}$ = [ $\frac{(n-1) \times s^{2} }{32.67 }$ , $\frac{(n-1) \times s^{2} }{11.59 }$ ]

= [ $\frac{21 \times 5.063^{2} }{32.67 }$ , $\frac{21 \times 5.063^{2} }{11.59 }$ ]

= [16.48 , 46.45]

<u>90% confidence interval for</u> $\sigma$ = [ $\sqrt{16.48}$ , $\sqrt{46.45}$ ]

= [4.06 , 6.82]

Therefore, a 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].

5 0

3 years ago