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erastova [34]
3 years ago
12

Which equation is represented by the graph below?

Mathematics
1 answer:
Stels [109]3 years ago
7 0

Answer:

y= - 1/2 +2

Step-by-step explanation:

One, the slope is negative, so you can cross some answers out right off the bat. Other than that, the answer is not a whole because it's not that steep. If you use RISE/RUN you can tell its a -1/2. Plus, the y-intercept is 2!

i hope this helps!!

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What is 0.329 in expanded form?
andrey2020 [161]
3/10 + 2/100 + 9/1000

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3 years ago
Help please______________________________
sesenic [268]

Answer:

By rounding each number to the nearst whole number, estimate the solution:

2.7²X4.3=31.347

8.8-5.15=3.65

for this 31 and 4 if you want to round

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
∠A and \angle B∠B are complementary angles. If m\angle A=(4x+24)^{\circ}∠A=(4x+24) ∘ and m\angle B=(3x-4)^{\circ}∠B=(3x−4) ∘ , t
Ostrovityanka [42]

Answer:

<h3>             m∠B = 26°</h3>

Step-by-step explanation:

∠A and ∠B being complementary means  m∠A + m∠B = 90°

4x + 24 + 3x - 4 = 90

7x + 20 = 90

  7x = 70

   x = 10

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7 0
4 years ago
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.

7 0
3 years ago
U GET BRAINLIEST FOR ANSWERING THIS RIGHT
Paul [167]

the answer is the second choice

5 0
4 years ago
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