All categories

2 years ago

Does the point (5 6) satisfy the equation y = 2x - 7

Mathematics

1 answer:

Aleks04 [339]2 years ago

5 0

Answer: No

Step-by-step explanation: To determine this, all we have to is plug our ordered pair into the equation for x and y.

Remeber that the first number is our x-coordinate and our second

number is our y-coordinate so we plug a 5 in for x and a 6 in for y.

So we have (6) = 2(5) + 7 or 6 = 10 + 7 or 6 = 17.

Since this is not a true statement, this point doesn’t satisfy the equation.

You might be interested in

To celebrate the first day of winter, Lin is putting magnets on some of the lockers (pre-COV ID of course). There are 100 locker

NikAS [45]

Two lockers, 35 and 70

7 0

2 years ago

Read 2 more answers

rusak2 [61]

Answer:

From top to bottom;

1,1,3,3

Step-by-step explanation:

mathematically, for an even function;

f(x) = f(-x)

what this mean is that;

f(-1) = f(1)

f(-3) = f(3)

f(-5) = f(5)

f(-6) = f(6)

so we have it that;

f(-1) = 1

f(-3) = 1

f(-5) = 3

f(-7) = 3

7 0

2 years ago

Helpp me asapp with this problem plss will crown brainiest !!!!

Evgen [1.6K]

Answer:

see below

Step-by-step explanation:

Since RT bisects QRS

1/2 QRS = TRS

1/2 ( 9x+214) = -9x+53

Multiply each side by 2

9x +214 = -18x+ 106

Add 18x from each side

9x +214+18x = 106

27x +214 = 106

Subtract 214 from each side

27x = 106-214

27x =-108

Divide by 27

27x/27 = -108/27

x = -4

TRS = -9x+53

=-9 *-4 +53

= 36+ 53

=89

4 0

3 years ago

Read 2 more answers

If the Semi Monthly is $4,000

Andreyy89

Disclaimer- this is all assuming that "semi monthly" means twice a month. if it doesn't, ignore this answer

A- if semi monthly is half a month, then 4 times 2 is 8, and 8 times 12 is 96,000.

B- If there are 4 weeks in a month and 4000$ accounts for half of that pay, then the weekly pay is 2000, making the bi-weekly pay 1000.

C- The monthly pay is 8,000$ because 4,000 x 2 is 8,000

D- The weekly pay is 2,000$ because the monthly pay is 8,000$ and there are 4 weeks in a month

E- I do not know the hours at this job and therefore cannot answer this

4 0

2 years ago

J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.

melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean for mail-order sales = $74.50

$\bar X_2$ = sample mean for internet sales = $84.40

$s_1$ = sample standard deviation for mail-order purchases = $17.25

$s_2$ = sample standard deviation for internet purchases = $21.25

$n_1$ = sample of sales receipts for mail-order purchases = 16

$n_2$ = sample of sales receipts for internet purchases = 9

Also, $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$ = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ( $\mu_1-\mu_2$ ).

Now, 99% confidence interval for ( $\mu_1-\mu_2$ ) is given by;

= $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

= $(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$

= [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0

3 years ago