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Mekhanik [1.2K]
3 years ago
13

Tina is saving to buy a notebook computer. She has two options. The first option is to put $300 away

Mathematics
1 answer:
KIM [24]3 years ago
3 0

Answer:

540

Step-by-step explanation:

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HI HELP ME W MATH PLSS!!
4vir4ik [10]

Answer:

a) 79.6

b) 437.8

Step-by-step explanation:

First, you have to do 318.4/4 to find the rate she goes in the duration of an hour. The rate is 79.6, which means she travels at a rate of 79.6 kilometers per hour.

Then, you have to find the distance she takes to travel in 5.5 hours, so you have to do 5.5*79.6 which is 437.8.

So, Hannah travels 437.8 kilometers in 5.5 hours.

6 0
3 years ago
The equation of line f is y = 1/4x -1. line g includes the point (1,-6) and is perpendicular to line f . What is the equation of
Gwar [14]
Y=-4x+b
-6=-4+b
-2=b
y=-4x-2
7 0
3 years ago
9) Simplify 2x^5-6^2+4x^4y/2x
KonstantinChe [14]

Answer:

x^4+2x^3y-3x

Step-by-step explanation:

6 0
3 years ago
Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in
Law Incorporation [45]

Answer:

(a) 900

(b) [567.35 , 1689.72]

(c) [23.82 , 41.11]

Step-by-step explanation:

We are given that a sample of 20 days of operation shows a sample mean of 290 rooms occupied per  day and a sample standard deviation of 30 rooms i.e.;

Sample mean, xbar = 290      Sample standard deviation, s = 30  and  Sample size, n = 20

(a) Point estimate of the population variance is equal to sample variance, which is the square of Sample standard deviation ;

                         \sigma^{2}  =  s^{2} = 30^{2}

                          \sigma^{2}  = 900

(b) 90% confidence interval estimate of the population variance is given by the pivotal quantity of  \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2} __n_-_1

P(10.12 < \chi^{2}__1_9 < 30.14) = 0.90 {At 10% significance level chi square has critical

                                           values of 10.12 and 30.14 at 19 degree of freedom}        

P(10.12 < \frac{(n-1)s^{2} }{\sigma^{2} } < 30.14) = 0.90

P(\frac{10.12}{(n-1)s^{2} } < \frac{1 }{\sigma^{2} } < \frac{30.14}{(n-1)s^{2} } ) = 0.90

P(\frac{(n-1)s^{2} }{30.14} < \sigma^{2} < \frac{(n-1)s^{2} }{10.12} ) = 0.90

90% confidence interval for \sigma^{2} = [\frac{19s^{2} }{30.14} , \frac{19s^{2} }{10.12}]

                                                   = [\frac{19*900 }{30.14} , \frac{19*900 }{10.12}]

                                                   = [567.35 , 1689.72]

Therefore, 90% confidence interval estimate of the population variance is [567.35 , 1689.72] .

(c) 90% confidence interval estimate of the population standard deviation is given by ;

       P(\sqrt{\frac{(n-1)s^{2} }{30.14}} < \sigma < \sqrt{\frac{(n-1)s^{2} }{10.12}} ) = 0.90

90% confidence interval for \sigma = [\sqrt{\frac{19s^{2} }{30.14}}   , \sqrt{\frac{19s^{2} }{10.12}}  ]

                                                 = [23.82 , 41.11]

Therefore, 90% confidence interval estimate of the population standard deviation is [23.82 , 41.11] .

7 0
3 years ago
Need help with sisters hw
tigry1 [53]

1426

× 357

________

509082

7 0
3 years ago
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