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ladessa [460]
3 years ago
8

An airplane 30,100 feet above the ground begins descending at the rate of 2150 feet per minute. Write and solve a linear equatio

n to find how long it will take the plane to reach the ground.
Mathematics
1 answer:
Inga [223]3 years ago
6 0

Answer:

14 minutes.

Step-by-step explanation:

To create a linear equation that represents an airplane flying at an altitude of 30,100 feet, descending 2,150 feet per minute, that represents how long it will take for the airplane to touch down, proceed as follows:

2,150X = 30,100

X = 30,100 / 2,150

X = 14

Thus, the plane will take 14 minutes to reach the ground, descending 2,150 feet per minute from 30,100 feet.

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Velvet fabric costs $1.1.49 a yard. Yvette buys 9.3 yards. Yvette's estimated cost of the velvet is
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This is a simple multiplcation and unit converstion problem. If 1 yard costs $1.149 and he gets 9.3 yards, you just multiply:
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6 0
3 years ago
Round 533,736 to the nearest thousand
tresset_1 [31]
The thousand if this number is 533,736.
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3 years ago
Choose all the functions that are NOT linear.
Triss [41]
Answer:
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8 0
3 years ago
An automobile company wants to determine the average amount of time it takes a machine to assemble a car. A sample of 40 times y
aksik [14]

Answer:

A 98% confidence interval for the mean assembly time is [21.34, 26.49] .

Step-by-step explanation:

We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                               P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average time = 23.92 minutes

             s = sample standard deviation = 6.72 minutes

             n = sample of times = 40

             \mu = population mean assembly time

<em> Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

<u>So, a 98% confidence interval for the population mean, </u>\mu<u> is; </u>

P(-2.426 < t_3_9 < 2.426) = 0.98  {As the critical value of z at 1%  level

                                               of significance are -2.426 & 2.426}  

P(-2.426 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.426) = 0.98

P( -2.426 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.426 \times {\frac{s}{\sqrt{n} } } ) = 0.98

P( \bar X-2.426 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.426 \times {\frac{s}{\sqrt{n} } } ) = 0.98

<u>98% confidence interval for</u> \mu = [ \bar X-2.426 \times {\frac{s}{\sqrt{n} } } , \bar X+2.426 \times {\frac{s}{\sqrt{n} } } ]

                                     = [ 23.92-2.426 \times {\frac{6.72}{\sqrt{40} } } , 23.92+2.426 \times {\frac{6.72}{\sqrt{40} } } ]  

                                    = [21.34, 26.49]

Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

7 0
3 years ago
Not easy help me please
Yuki888 [10]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

In the number line attached in my attachment,

The red line represents point :

  • \dfrac{10}{2}

The blue line represents point :

  • \dfrac{ - 9}{2}

I hope it helps ~

3 0
3 years ago
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