The equation y = [x

Punctele O, A, B, C sunt coliniare în această ordine, iar segmentele OA, AB și BC au lungimile exprimate în cm, prin numere natu

julia-pushkina [17]

Răspuns:

a) OC = 18cm

b) MN = 6cm

Explicație pas cu pas:

Găsiți diagrama la întrebarea atașată.

a) Din diagramă se poate observa că OC = OA + AB + AC

OC = 2n + 2n + 2 + 2n + 4

OC = 6n + 6

Următorul este să obțineți valoarea lui n:

OM = OA + AM

Să obținem segmentul AM;

Deoarece M este segmentul mijlociu al AC, atunci;

AM = AC / 2

AM = AB + BC / 2

AM = 2n + 2 + 2n + 4/2

AM = 4n + 6/2

AM = 2n + 3

Din moment ce OM = OA + AM

OM = 2n + 2n + 3

OM = 4n + 3

Dat fiind OM = 11

11 = 4n + 3

4n = 11-3

4n = 8

n = 2

Următorul pas este să calculați OC;

OC = 6n + 6

OC = 6 (2) + 6

OC = 18

Prin urmare, lungimea segmentului OC este de 18 cm

b) Această întrebare ne cere să găsim segmentul de linie MN.

Din diagramă, MN = AM - AN

Deoarece AM = 2n + 3 din întrebarea 1;

AM = 2 (2) +3

AM = 4 + 3

AM = 7

Acum, să obținem AN:

Din diagramă, AN = ON-OA și ON = OB / 2

AN = OB / 2 - OA

AN = (2n + 2n + 2) / 2 - 2n

AN = 4n + 2/2 - 2n

AN = 2n + 1 - 2n

AN = 2n-2n + 1

AN = 1

De la MN = AM - AN

MN = 7 - 1

MN = 6cm

Prin urmare, distanța dintre N, mijlocul segmentului OB și punctul M este de 6cm

6 0

3 years ago

The balance on a credit card, that charges a 15.5%APR interest rate, over a 1 month period is given inthe following table:Days 1

Karo-lina-s [1.5K]

First step is to get the average daily balance.

Take the sum of each day's balances.

If a payment has been made, the sign will be negative.

From Days 1 - 5 :

Day 1 = $200

Day 2 = $200

...

Day 5 = $200

That's $200 x 5 days = $1000

From Days 6 - 20 :

Day 6 = $350

Day 7 = $350

...

Days 20 = $350

That's $350 x 15 days = $5250

From Days 21 - 30 :

Day 21 = $150

Day 22 = $150

...

Day 30 = $150

That's $150 x 10 days = $1500

A total of :

$1000 + $5250 + $1500 = $7750

Now divide this total to the number of days to get the ADB or Average Daily Balance.

$\text{ADB}=\frac{7750}{30}=258.33$

Next Step is to calculate for the finance charge using the formula :

$\text{Charge}=\text{ADB}\times r\times d$

where ADB is the average daily balance

r is the interest rate per day

d is the number of days or period given

From the problem, APR is 15.5%, it means that we need to divide the APR by 365 days.

So r = 0.155/365 = 0.000425

And we have d = 30 days

The charge will be :

Charge = 258.33 x 0.000425 x 30 = $3.29

The answer is $3.29

5 0

1 year ago

For which equation is s = 9 not the solution?

svlad2 [7]

Answer:

what is the equations?

3 0

2 years ago

7. Jenny is saving to buy a laptop worth $960. Her parents promise to match every $3 saved

PSYCHO15rus [73]

First off, if you divide $960/$8, you will get 120 “sets” of payment. Multiply this value by the amount of money Jenny is responsible for paying and the amount her parents are responsible for paying to find their total contributions.

a. Jenny will have to pay $600

b. Jenny’s parents will contribute $360 to her new laptop.

8 0

2 years ago

Listed below are the lead concentrations in mu ??g/g measured in different traditional medicines. Use a 0.10 0.10 significance l

laiz [17]

Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

We are given below are the lead concentrations in mu;

6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20

We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.

Let $\mu$ = mean lead concentration for all such medicines.

So, Null Hypothesis, $H_0$ : $\mu$ = 17 mu {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, $H_A$ : $\mu$ < 17 mu {means that the mean lead concentration for all such medicines is less than 17 mu}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}$ ~ $t_n_-_1$

where, $\bar X$ = sample mean lead concentration = $\frac{\sum X}{n}$ = 12.25 mu

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 6.96 mu

n = sample size = 10

So, test statistics = $\frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}$ ~ $t_9$

= -2.16

The value of t test statistics is -2.16.

Now, the P-value of the test statistics is given by the following formula;

P-value = P( $t_9$ < -2.16) = 0.031

Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test. Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.

7 0

3 years ago

The equation y = [x - 5] +2 is?