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2 years ago

What is y=3(2x+1)+5 work need to be shown

Mathematics

1 answer:

jonny [76]2 years ago

6 0

Answer:

In slope intercept form, this equation is y = 6x + 8

Step-by-step explanation:

All you need to do to get this to slope intercept form is to distribute the term and then combine like terms.

y = 3(2x + 1) + 5 ----> Distribute

y = 6x + 3 + 5 ----> Combine like terms

y = 6x + 8

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shayla is financing a house for $126,750. She has to pay $450 plus 1.15% for a brokerage fee. How much are the mortgage brokerag

ANEK [815]

Answer: $1,907.63

Explanation:

It is stated in the problem that the brokerage fee is $450 plus 1.15% (meaning 1.15% of $126,750). Hence the brokerage fee is computed as follows

(Brokerage fee) = $450 + (1.15% of $126,750)
= $450 + (0.0115)($126,750)
= $450 + $1,457.625
= $1,907.625

Since there is no half cents today, we round-off the brokerage fee to the nearest cent. Hence the brokerage fee is $1,907.63.

Note: In the computation of brokerage fee, we need to change 1.15% to decimal.

7 0

2 years ago

A detergent company periodically tests its products for variations in the fill weight. To do this, the company uses x-bar and R

maw [93]

Answer:

10.9361

Step-by-step explanation:

The lower control limit for xbar chart is

xdoublebar-A2(Rbar)

We are given that A2=0.308.

xdoublebar=sumxbar/k

Rbar=sumR/k

xbar R

5.8 0.42

6.1 0.38

16.02 0.08

15.95 0.15

16.12 0.42

6.18 0.23

5.87 0.36

16.2 0.4

Xdoublebar=(5.8+6.1+16.02+15.95+16.12+6.18+5.87+16.2)/8

Xdoublebar=88.24/8

Xdoublebar=11.03

Rbar=(0.42+0.38+0.08+0.15+0.42+0.23+0.36+0.4)/8

Rbar=2.44/8

Rbar=0.305

The lower control limit for the x-bar chart is

LCL=xdoublebar-A2(Rbar)

LCL=11.03-0.308*0.305

LCL=11.03-0.0939

LCL=10.9361

8 0

3 years ago

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

$A=\sqrt{s(s-a)(s-b)(s-c)}$ , where $a$ , $b$ , and $c$ are three sides of the triangle and $s$ is the semi-perimeter ( $s=\frac{a+b+c}{2}$ ).

The semi-perimeter, $s$ , is:

$s=\frac{5+5+4}{2}=\frac{14}{2}=7$

Therefore, the area of the isosceles triangle is:

$A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}$

Thus, the area of the quadrilateral is:

$6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}$

4 0

3 years ago

WHO KNOWS THIS? I NEED HELP PLS

NeX [460]

Answer:

sorry i do not know

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

A multiple-choice examination has 20 questions, each with five possible answers, only one of which is correct. Suppose that one

kari74 [83]

Answer:

The probability that the student answers at least seventeen questions correctly is $8.03\times 10^{-10}$ .

Step-by-step explanation:

Let the random variable X represent the number of correctly answered questions.

It is provided all the questions have five options with only one correct option.

Then the probability of selecting the correct option is,

$P(X)=p=\frac{1}{5}=0.20$

There are n = 20 question in the exam.

It is also provided that a student taking the examination answers each of the questions with an independent random guess.

Then the random variable can be modeled by the Binomial distribution with parameters n = 20 and p = 0.20.

The probability mass function of X is:

$P(X=x)={20\choose x}\ 0.20^{x}\ (1-0.20)^{20-x};\ x =0,1,2,3...$

Compute the probability that the student answers at least seventeen questions correctly as follows:

$P(X\geq 17)=P (X=17)+P (X=18)+P (X=19)+P (X=20)$

$=\sum\limits^{20}_{x=17}{{20\choose x}\ 0.20^{x}\ (1-0.20)^{20-x}}\\\\=0.00000000077+0.000000000032+0.00000000000084+0.000000000000042\\\\=0.000000000802882\\\\=8.03\times10^{-10}$

Thus, the probability that the student answers at least seventeen questions correctly is $8.03\times 10^{-10}$ .

4 0

3 years ago