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3 years ago

PLEASE ANSWER I WILL GIVE YOU BRAINIEST IF CORRECT!!!!!!! thanks!!

Mathematics

1 answer:

xeze [42]3 years ago

4 0

Answer:

100*5(120) = $60,000 for 120 students

Step-by-step explanation:

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To answer that question

Mars2501 [29]

First, we get ax^2+bx+c. Next, we know that the line of symmetry is -b/2a. Since we know that there is a maximum value, the parabola is facing downwards, so a is negative. For random numbers, we can say that a = -0.5 and b=-10 (b needs to be negative for -b/2a to equal -10), getting -0.5x^2-10x+c. Plugging -10 in for x (since -10 is the middle it is the max), we get -50+100=50. Since the maximum needs to be 5, not 50, we subtract 45 from the answer to get it and therefore make c = -45, getting -0.5x^2-10x-45

5 0

3 years ago

How to pass my regions

Anettt [7]

by studying because that's the hard truth

4 0

3 years ago

Read 2 more answers

Four different paints are advertised as having the same drying time. To check the manufacturer's claims, five samples were teste

iogann1982 [59]

Answer:

Explained below.

Step-by-step explanation:

The data provided is for the dying time of four different types of paint.

One-way ANOVA can be used to determine whether all the four paints have the same drying time.

Use Excel to perform the one-way ANOVA.

Go to Data → Data Analysis → Anova: Single Factor

A dialog box will open.

Select the data.

Select "Grouping" as Columns.

Press OK.

The output is attached below.

The required values are as follows:

(1)

Sum of Squares of Treatment (Between Subjects):

SST = 330

(2)

Sum of Squares of Error (Within Subjects):

SSE = 692

(3)

Mean Squares Treatment (Between Subjects):

MST = 110

(4)

Mean Squares Error (Within Subjects):

MSE = 43.25

6 0

3 years ago

Of 560 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compared

frutty [35]

Answer:

a) (0.5256,0.5944)

c) Criticism is invalid

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 560

Proportion of mislabeled = 56%

$\hat{p} = 0.56$

a) 90% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.64$

Putting the values, we get:

$0.56\pm 1.64(\sqrt{\dfrac{0.56(1-0.56)}{560}}) = 0.56\pm 0.0344\\\\=(0.5256,0.5944)$

b) Interpretation of confidence interval:

We are 90% confident that the true proportion of all seafood in the country that is mislabeled or misidentified is between 0.5256 and 0.5944 that is 52.56% and 59.44%.

c) Validity of criticism

Conditions for validity:

$np > 10\\n(1-p)>10$

Verification:

$560\times 0.56 = 313.6>10\\560(1-0.56) = 246.4>10$

Both the conditions are satisfied. This, the criticism is invalid.

8 0

4 years ago

Jbk111 only PLEASE<br> 27+2,993=??

m_a_m_a [10]

3,020

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7 0

3 years ago