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2 years ago

5

A professional writer decides to judge the value of her laptop by the number of times she presses

1 answer:

shusha [124]2 years ago

7 0

Answer:

D.)

15040000 <= k <=19840000

Step-by-step explanation:

V = 1200 -0.00002k

Value V after one year falls in the range of $803.20 and $899.20.

Let's substract these values from 1200.

1200-803.20=396.80

1200-899.20=300.80

So it means that

0.00002k= 396.80

And

0.00002k= 300.80

So

0.00002k= 396.80

K= 396.80/0.00002

K= 19840000

And

0.00002k= 300.80

K = 300.80/0.00002

K= 15040000

So range is between

15040000 <= k <=19840000

You might be interested in

To make f continuous at x=2 , f(2) should be defined as what value? Justify your answer.

yawa3891 [41]

Answer:

To make f continuous at $x=2$ , $f(2)$ should be defined $x = 2.$

Step-by-step explanation:

A function, let say $f(x)$ , is defined at $x = c$ is continuous at $x = c$

If the limit of $f(x)$ as $x$ approaches $c$ is equal to the value of $f(x)$ at $x = c$ .

Mathematically it is written as:

if

$\lim _{x\to c}\:f\left(x\right)=f\left(c\right)$

then

$f(x)$ is continuous at $x = c$ .

So from the above definition, we conclude that:

To make f continuous at $x=2$ , $f(2)$ should be defined $x = 2.$

i.e.

if

$\lim _{x\to 2}\:f\left(x\right)=f\left(2\right)$

then

$f(x)$ is continuous at $x = 2$ .

8 0

1 year ago

4. Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln

pochemuha

Answer:

No these these result do not differ at 95% confidence level

Step-by-step explanation:

From the question we are told that

The first concentrations is $c _1= 30.0 \ g/m^3$

The second concentrations is $c _2 = 52.9 \ g/m^3$

The first sample size is $n_1 = 32$

The second sample size is $n_2 = 32$

The first standard deviation is $\sigma_1 = 30.0$

The first standard deviation is $\sigma_1 = 29.0$

The mean for Turnpike is $\= x _1 = \frac{c_1}{n} = \frac{31.4}{32} = 0.98125$

The mean for Tunnel is $\= x _2 = \frac{c_2}{n} = \frac{52.9}{32} = 1.6531$

The null hypothesis is $H_o : \mu _1 - \mu_2 = 0$

The alternative hypothesis is $H_a : \mu _1 - \mu_2 \ne 0$

Generally the test statistics is mathematically represented as

$t = \frac{\= x_1 - \= x_2}{ \sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }}$

$t = \frac{0.98125 - 1.6531}{ \sqrt{\frac{30^2}{32} +\frac{29^2}{32} }}$

$t = - 0.0899$

Generally the degree of freedom is mathematically represented as

$df = 32+ 32 - 2$

$df = 62$

The significance $\alpha$ is evaluated as

$\alpha = (C - 100 )\%$

=> $\alpha = (95 - 100 )\%$

=> $\alpha =0.05$

The critical value is evaluated as

$t_c = 2 * t_{0.05 , 62}$

From the student t- distribution table

$t_{0.05, 62} = 1.67$

So

$t_c = 2 * 1.67$

=> $t_c = 3.34$

given that

$t_c > t$ we fail to reject the null hypothesis so this mean that the result do not differ

6 0

1 year ago

I need help with number 5 please

steposvetlana [31]

Answer: Would be 55% I hope this helps!

7 0

2 years ago

Read 2 more answers

Determine whether or not the vector field is conservative. if it is conservative, find a function f such that f = ∇f. (if the ve

sweet [91]

A conservative vector field $\mathbf f$ has curl $\nabla\times\mathbf f=\mathbf0$ . In this case,

$\nabla\times\mathbf f=12xy^2z(1-z)\,\mathbf j+4yz^2(2z-3x)\,\mathbf k\neq\mathbf 0$

so the vector field is not conservative.

3 0

2 years ago

At time t=3, there are 2.5 cubic feet of water in the tub. Write an equation for the locally linear approximation of W at t=3, a

velikii [3]

Answer:

W=0.8333*t

2.917 ft³

Step-by-step explanation:

At time t=3, there is 2.5 ft³ of water in the tub

Finding the relation will be;

3=2.5

1=?

Rate=0.8333 ft³/min

W=0.8333 ft³ / min

W=0.8333*t

At t=3.5 will be;

1 min = 0.8333

3.5 min =?

Apply cross-production

3.5*0.8333 =2.917 ft³

4 0

2 years ago

Read 2 more answers