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never [62]
2 years ago
6

0.2(x + 15) = 0.5(3x - 7)

Mathematics
1 answer:
Dafna1 [17]2 years ago
8 0
Are you asking if this is greater than, less than, or equal to?
You might be interested in
Approximately 52% of all recent births were boys. In a simple random sample of 100 recent births, 49 were boys and 51 were girls
Leni [432]

Answer:

Option A is right

Step-by-step explanation:

Given that approximately 52% of all recent births were boys. In a simple random sample of 100 recent births, 49 were boys and 51 were girls. The most likely explanation for the difference between the observed results and the expected results in this case is

A) variability due to sampling

-- True because there is a slight difference whichmay be due to sampling fluctuations.

B) bias

False because given that 100 random births selected

C) nonsampling error

False.  There is no chance for systematic error here.

d) Confounding:  There is no confounding variable present inchild birth since each is independent of the other

e) a sampling frame that is incomplete

False because the sampling is done correctly.

8 0
2 years ago
Solve for y.<br> -38 = 7v+17 – 12v
iren2701 [21]

Answer:

<h2>v = 11</h2>

Step-by-step explanation:

-38=7v+17-12v\qquad\text{subtract 17 from both sides}\\\\-38-17=7v+17-17-12v\qquad\text{combine like terms}\\\\-55=-5v\qquad\text{divide both sides by (-5)}\\\\\dfrac{-55}{-5}=\dfrac{-5v}{-5}\\\\11=v\to v=11

6 0
3 years ago
You have two squares. The larger square has a side of 3 more than the smaller square. If the combined area of the two squares is
joja [24]
The length of the SMALLER square is:
B. 5

Basically I tested each answer choice and see if all the information matches it.
Lets check how my answer is correct:
We know that:
The length of the bigger square is 3 more than the smaller square’s length.
The areas of both squares add up to 89 cm^2
length of the SMALLER square is: 5 cm
length of the LARGER square:
5 + 3 = 8 cm

Area of smaller square:
5^2 = 25 cm^2

Area of bigger square:
8^2 = 64 cm^2

ADD UP BOTH AREAS OF SQUARES:
25 + 64 = 89 cm^2

Hope this helps!
3 0
2 years ago
Read 2 more answers
Joanna is driving 185 miles to visit her family. She drives 68 miles in the morning, stops for lunch, and then drives the rest o
Nuetrik [128]

Answer:

68+m=185

m=117

Step-by-step explanation:

68 miles before lunch, plus m miles after lunch is a total of 185 miles. Assuming she reaches her destination. That creates the equation 68+m=185

Then, to figure out the miles she travels after lunch we solve for m:

do this by subtracting 68 from both sides. This leaves us with: m=117

5 0
3 years ago
Find the perimeter and the area of the figure.
Gwar [14]

\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}

<u>the </u><u>given </u><u>figure </u><u>is </u><u>a </u><u>composition</u><u> </u><u>of </u><u>a </u><u>rectangle</u><u> </u><u>as </u><u>well </u><u>as </u><u>a </u><u>right </u><u>angled </u><u>triangle </u><u>!</u>

<u>we've</u><u> </u><u>been </u><u>given </u><u>the </u><u>two </u><u>sides </u><u>of </u><u>the </u><u>rectangle </u><u>and </u><u>we're</u><u> </u><u>required</u><u> </u><u>to </u><u>find </u><u>out </u><u>the </u><u>height </u><u>of </u><u>the </u><u>triangle </u><u>,</u><u> </u><u>so </u><u>as </u><u>to </u><u>find </u><u>it's</u><u> </u><u>area </u><u>~</u>

<u>we </u><u>know </u><u>the </u><u>the </u><u>opposite</u><u> </u><u>sides </u><u>of </u><u>a </u><u>rectangle </u><u>are </u><u>equal</u><u> </u><u>,</u><u> </u><u>therefore </u><u>we </u><u>can </u><u>break </u><u>the </u><u>longest </u><u>side </u><u>(</u><u> </u><u>length </u><u>=</u><u> </u><u>9</u><u>.</u><u>5</u><u> </u><u>cm </u><u>)</u><u> </u><u>into </u><u>two </u><u>parts </u><u>!</u><u> </u><u>the </u><u>first </u><u>part </u><u>of </u><u>length </u><u>=</u><u> </u><u>7</u><u> </u><u>cm </u><u>which </u><u>is </u><u>the </u><u>length </u><u>of </u><u>the </u><u>rectangle </u><u>and </u><u>the </u><u>rest </u><u>2</u><u>.</u><u>5</u><u> </u><u>cm </u><u>(</u><u> </u><u>9</u><u>.</u><u>5</u><u> </u><u>-</u><u> </u><u>7</u><u> </u><u>=</u><u> </u><u>2</u><u>.</u><u>5</u><u> </u><u>)</u><u> </u><u>will </u><u>become </u><u>the </u><u>height </u><u>of </u><u>the </u><u>triangle </u><u>!</u>

<h3><u>For </u><u>perimeter</u><u> </u><u>of </u><u>the </u><u>figure </u><u>-</u></h3>

perimeter \: of \: figure = perimeter \: of \: rectangle + perimeter \: of \: triangle \\  \\

now ,

<u>perimeter</u><u> </u><u>of </u><u>rectangle </u><u>=</u><u> </u><u>2</u><u> </u><u>(</u><u> </u><u>l </u><u>+</u><u> </u><u>b </u><u>)</u>

where ,

<u>l </u><u>=</u><u> </u><u>length </u>

<u>b </u><u>=</u><u> </u><u>breadth </u>

\longrightarrow \: perimeter = 2(7 + 6) \\ \longrightarrow \: 2(13) \\ \longrightarrow \: 26 \: cm

and ,

perimeter \: of \: \triangle = 6.5 + 2.5 + 6 \\ \longrightarrow \: 15 \: cm

<u>Perimeter</u><u> </u><u>of </u><u>figure </u><u>in </u><u>total </u><u>=</u><u> </u><u>2</u><u>6</u><u> </u><u>cm </u><u>+</u><u> </u><u>1</u><u>5</u><u> </u><u>cm</u>

thus ,

\qquad\quad\bold\red{perimeter \: = \: 41 \: cm}

<h3><u>For </u><u>area </u><u>of </u><u>the </u><u>figure </u><u>-</u></h3>

area \: of \: figure = area \: of \: rectangle  + area \: of \: rectangle \\

now ,

<u>area </u><u>of </u><u>rectangle</u><u> </u><u>=</u><u> </u><u>l </u><u>×</u><u> </u><u>b</u>

where ,

<u>l </u><u>=</u><u> </u><u>length </u>

<u>b </u><u>=</u><u> </u><u>breadth</u>

area \: of \: rectangle = 7 \times 6 \\ \longrightarrow \: 42 \: cm {}^{2}

and ,

area \: of\triangle =  \frac{1}{2}  \times base \times height \\  \\ \longrightarrow \:   \frac{1}{\cancel2}  \times \cancel6 \times 2.5 \\  \\ \longrightarrow \: 3 \times 2.5 \\  \\ \longrightarrow \: 7.5 \: cm {}^{2}

<u>Area </u><u>of </u><u>figure</u><u> </u><u>in </u><u>total </u><u>=</u><u> </u><u>4</u><u>2</u><u> </u><u>cm²</u><u> </u><u>+</u><u> </u><u>7</u><u>.</u><u>5</u><u> </u><u>cm²</u>

thus ,

\qquad\quad\bold\red{Area \: = \: 49.5 \: cm^{2}}

hope helpful :)

6 0
1 year ago
Read 2 more answers
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