Ask question

Ask question

All categories

3 years ago

6

The equation y = 1725 - 75x represents the amount of money y (in dollars) left in your lunch account after x weeks. 20 weeks go

by. FInd the domain and range of the function.

1 answer:

garik1379 [7]3 years ago

6 0

Y= -75x +1725
y= -75(20) +1725
y = -1500 + 1725
y= 225 = range
225= -75x +1725
225 - 1725= -75x
-1500=-75x
-1500/-75 = -75x/-75
x = 20 = domain

You might be interested in

What is 15 tens - 1 gross

vampirchik [111]

Answer:

see the explanation

Step-by-step explanation:

we know that

A gross is equal to 120 ones or ten dozen

what is 15 tens - 1 gross

we know that

15 tens means ----> That you are adding 10, 15 times or multiplying 10 by 15, which gives you

$10(15)=150$

1 gross means ---> That you are adding 10, 12 times or multiplying 10 by 12

which gives you

$10(12)=120$

so

The algebraic expression of 15 tens - 1 gross is equal to

$150-120=30$

Convert to word expression

3 tens

8 0

3 years ago

The following frequency table relates the weekly sales of bicycles at a given store over a 42-week period.

7nadin3 [17]

Answer:

14

Step-by-step explanation:

6 0

3 years ago

Find the missing lengths:<br> LO=5 and OK=4, find OH and KH.

hodyreva [135]

Answer:

The length of KH is 6 units and OH is 6.3 units.

Step-by-step explanation:

Given the figure with lengths LO=5 and OK=4. we have to find the length of OH and KH.

In ΔLOH

By Pythagoras theorem

$LH^2=LO^2+OH^2\\\\LH^2=5^2+OH^2$ → (1)

In ΔKOH,

$KH^2=OH^2+OK^2\\\\KH^2=OH^2+4^2$ → (2)

In ΔKHL,

$KL^2=LH^2+KH^2$

Using eq (1) and (2), we get

$KL^2=5^2+OH^2+OH^2+4^2$

$9^2=25+2OH^2+16$

⇒ $2OH^2=81-25-16=40$

⇒ $OH=\sqrt{20}=6.324\sim6.3units.$

Put the above value in eq 2, we get

$KH^2=20+4^2=36$

⇒ KH=6 units.

8 0

3 years ago

Read 2 more answers

(+7) + (+5) = *<br> Nour aſiswer

earnstyle [38]

Answer:

12

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A university wants to compare out-of-state applicants' mean SAT math scores (?1) to in-state applicants' mean SAT math scores (?

nordsb [41]

Answer:

d. Yes, because the confidence interval does not contain zero.

Step-by-step explanation:

We are given that the university looks at 35 in-state applicants and 35 out-of-state applicants. The mean SAT math score for in-state applicants was 540, with a standard deviation of 20.

The mean SAT math score for out-of-state applicants was 555, with a standard deviation of 25.

Firstly, the Pivotal quantity for 95% confidence interval for the difference between the population means is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean SAT math score for in-state applicants = 540

$\bar X_2$ = sample mean SAT math score for out-of-state applicants = 555

$s_1$ = sample standard deviation for in-state applicants = 20

$s_2$ = sample standard deviation for out-of-state applicants = 25

$n_1$ = sample of in-state applicants = 35

$n_2$ = sample of out-of-state applicants = 35

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 20^{2} +(35-1)\times 25^{2} }{35+35-2} }$ = 22.64

<em>Here for constructing 95% confidence interval we have used Two-sample t test statistics.</em>

So, 95% confidence interval for the difference between population means ( $\mu_1-\mu_2$ ) is ;

P(-1.997 < $t_6_8$ < 1.997) = 0.95 {As the critical value of t at 68 degree

of freedom are -1.997 & 1.997 with P = 2.5%}

P(-1.997 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 1.997) = 0.95

P( $-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.95

P( $(\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.95

<u>95% confidence interval for</u> ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

=[ $(540-555)-1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }$ , $(540-555)+1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }$ ]

= [-25.81 , -4.19]

Therefore, 95% confidence interval for the difference between population means SAT math score for in-state and out-of-state applicants is [-25.81 , -4.19].

This means that the mean SAT math scores for in-state students and out-of-state students differ because the confidence interval does not contain zero.

So, option d is correct as Yes, because the confidence interval does not contain zero.

6 0

3 years ago