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Wittaler [7]
3 years ago
11

Plz helppppp will add u as friend

Mathematics
2 answers:
Dafna11 [192]3 years ago
7 0
The answer is down below.
elixir [45]3 years ago
5 0

Answer:

y=1.44

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS
  • Equality Properties

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\frac{4.5}{y} =\frac{12.5}{4}<u />

<u />

<u>Step 2: Solve for </u><em><u>y</u></em>

  1. Cross-multiply:                    4(4.5) = 12.5y
  2. Multiply:                               18 = 12.5y
  3. Isolate <em>y</em>:                              1.44 = y
  4. Rewrite:                               y=1.44

<u>Step 3: Check</u>

<em>Plug in y to verify it's a solution.</em>

  1. Substitute:                    \frac{4.5}{1.44} =\frac{12.5}{4}
  2. Divide:                          3.125=3.125

Here, we see that 3.125 does indeed equal 3.125. ∴ y = 1.44 is a solution of the equation.

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Y=2−9x;x=2/3<br><br> just kinda stuck with fractions lol
Yuki888 [10]

Answer:

-4

Step-by-step explanation:

What I did was multiply 9/1 with 2/3 giving us 18/3 which simplifies into 6. Then what we do is subtract 2 with 6 giving us -4.

8 0
3 years ago
Read 2 more answers
A sample of 900 college freshmen were randomly selected for a national survey. Among the survey participants, 372 students were
Rufina [12.5K]

Answer:

a) ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423

b) 0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371

Step-by-step explanation:

1) Data given and notation  

n=900 represent the random sample taken    

X=372 represent the students were pursuing liberal arts degrees

\hat p=\frac{372}{900}=0.413 estimated proportion of students were pursuing liberal arts degrees

\alpha=0.01 represent the significance level

z would represent the statistic (variable of interest)    

p_v represent the p value (variable of interest)    

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

The margin of error is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

If we replace we have:

ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423

And replacing into the confidence interval formula we got:

0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371

0.413 + 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.455

And the 99% confidence interval would be given (0.371;0.455).

We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.

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3 years ago
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vladimir2022 [97]
Cohésion is the property
7 0
4 years ago
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Jack can run a mile in 9, 1/3 minutes. How long will it take for him to run 3, 1/2 miles?
BARSIC [14]
If you turn 9 1/3 in to a decimal and 3 1/2 in a decimal also 

9 1/3 = 9.3 and 3 1/2 =3.5 
9.3+3.5 =12.8 
12.8=12 4/5


8 0
4 years ago
Please help me with this question. Question C (i) and (ii)
GenaCL600 [577]

Answer:

idk

hahahahahaha

Step-by-step explanation:

8 0
3 years ago
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