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Dahasolnce [82]
3 years ago
5

Evaluate f(x)=4x+1 when x=−1, x=0, and x=3.

Mathematics
2 answers:
larisa86 [58]3 years ago
8 0

Answer:

when

Step-by-step explanation:

Dahasolnce [82]3 years ago
3 0

Answer:

-3, 1, and 13

Step-by-step explanation:

just plug in the numbers to the variable ex: 4(-1)+1

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Help me with this pls and ty
Bingel [31]

Answer:

It is the second answer.

Step-by-step explanation:

When numbers with the same base are multiplied, you add the bases.

1/2 + 3/4 is the same as

2/4 + 3/4 = 5/4

So now we have 2 ^5/4^2.  When we have a power raised to a power.  We multiply the powers.

5/4(2)

(5/4(2/1) = 10/4 = 5/2.  The second answer is the same way to describe

2^(5/2)

7 0
2 years ago
What is the picture equal to? Help please
amm1812

25(3n - 2) \\  \\ 1. \: 25  \times 3n + 25  \times  - 2 \\ 2. \: 75n + 25 \times  - 2 \\ 3. \: 75n - 50

7 0
3 years ago
Which is greater <br> 7/12 <br> 10/12
shtirl [24]
10/12 is more than 7/12


3 0
4 years ago
Read 2 more answers
A tomato plant grew an average of 0.06 m per day. How many days did it take for the plant to grow a total of 0.96 m.            
Nadya [2.5K]
If you divide 96 by 6 then you get 16. So it's 16 days!
3 0
4 years ago
Read 2 more answers
Two coworkers commute from the same building. They are interested in whether or not there is any variation in the time it takes
jenyasd209 [6]

Answer:

Step-by-step explanation:

Hello!

You have two samples.

Sample 1 (time that takes one commute of worker 1)

n₁= 20

S₁²= 12.2

Sample 2 (time that takes one commute of worker 2)

n₂= 20

S₂²= 16.9

The hypothesis is that the first coworker is more consistent with his commute times, this means that the variability of his commute times is less than the variability of the commute times of worker 2. With this in mind, the hypothesis for a variance ratio test is:

H₀: σ₁² ≥ σ₂²

H₁:: σ₁² < σ₂²

α: 0.10

The statistic for this test is:

F= (S₁²/S₂²) * (σ₁²/σ₂²) ~ F_{n1-1;n2-1}

The test is one-tailed (left) with just one critical value:

F_{n_1-1;n_2-1;\alpha } = \frac{1}{F_{n_2-1;n_1-1;1-\alpha }} = \frac{1}{F_{19;19;0.90}} = \frac{1}{1.82} =0.549

(The F-table I use only has the values for high probabilities so I've used a conversion method to obtain the value for the small probability)

Your decision rule is:

If F ≤ 0.549, you reject the null hypothesis.

if F > 0.549, you support the null hypothesis.

The calculated value:

F= (S₁²/S₂²) * (σ₁²/σ₂²) = (12.2/16.9) * 1= 0.722

Since the calculated F-value is greater than the critical value, the decision is to not reject the null hypothesis. So you can say that the variability of the commute times of worker 1 is less than the variability of the commute times of worker 2.

I hope you have a SUPER day!

7 0
3 years ago
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