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Ugo [173]
3 years ago
12

Brainly y=-1/3x+5 graph

Mathematics
1 answer:
Wewaii [24]3 years ago
8 0

Answer:

The answer will be listed below.

Step-by-step explanation:

To graph the equation, first mark down your y-intercept which would be 5. Look at the y-axis and put a dot on 5. Now you must graph the slope of the line. To graph the slope, you would use rise/run (1/3). You would start on 5 and go up one and to the right three times. Keep repeating the process until you run out. Now draw the line through the dots and you did it!

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Hewo Please Help THX!!! :)
Serga [27]

Answer:

$30

Step-by-step explanation:

x =4(3.45)+1.5(3.98)+2(4.35)+1.53

=13.8+5.97+8.7+1.53

=30

6 0
3 years ago
Allen�s hummingbird (Selasphorus sasin ) has been studied by zoologist Bill Alther A small group of 15 Allen� s hummingbirds has
Bogdan [553]

Answer:

1) 80% CI: [3.04; 3.26]gr

d= 0.11

2) n= 28 hummingbirds

Step-by-step explanation:

Hello!

The study variable of this experiment is:

X: the weight of a hummingbird. (gr)

And it has a normal distribution, symbolically: X~N(μ;σ²)

And (I hope I got it correctly) its population standard deviation is σ= 0.33

There was a sample of n= 15 hummingbirds taken, its sample mean X[bar]= 3.15 gr

1) You need to construct an 80% Confidence Interval for the population mean of the hummingbird's weight.

Since the study variable has a normal distribution, you can use either the standard normal distribution or the Student's t distribution. Both are useful to estimate the population mean. Since the population standard variance is known, the best choice is the Standard normal.

Z= <u> X[bar] - μ </u>~ N(0;1)

       σ/√n

The formula for the interval is:

X[bar] ± Z_{1- \alpha /2} * (σ/√n)

Z_{1- \alpha /2}= Z_{0.90} = 1.28

3.15 ± 1.28 * (0.33/√15)

[3.04; 3.26]gr

The margin of error (d) of a confidence interval is hal its amplitude (a)

a= Upper bond - Lower bond

d= (Upper bond - Lower bond)/2

d= \frac{(3.26-3.04)}{2} = 0.11

2) You need to calculate a sample size for a 80% Confidence interval for the average weight of the hummingbirds with a margin of error of d= 0.08

As I said before, the margin of error is half the amplitude of the interval, the formula you use to estimate the population mean has the following structure:

"point estamator" ± "margin of error"

Then the margin of error is:

d= Z_{1- \alpha /2} * (σ/√n)

Now what you have to do is rewrite the formula based on the sample size

d= Z_{1- \alpha /2} * (σ/√n)

\frac{d}{Z_{1- \alpha /2}}= σ/√n

√n * \frac{d}{Z_{1- \alpha /2}}= σ

√n = σ * \frac{Z_1- \alpha /2}{d}

n = (σ * \frac{Z_1- \alpha /2}{d})²

n=  (0.33 * \frac{1.28}{0.08})²

n= 27.8784 ≅ 28 hummingbirds.

I hope it helps!

4 0
3 years ago
Explain Conversions. Giving out easy questions, basically free points
Paul [167]

Answer:

the process of changing or causing something to change from one form to another.

similar to transformation

3 0
3 years ago
Read 2 more answers
Which statement is correct PLEASE HELP ILL GOVE YOU BRAINLIEST.
Valentin [98]
The first one is correct, just match the numbers with the corresponding sides of each triangle
6 0
3 years ago
Kelli swam upstream for some distance in one hour. She then swam downstream the same river for the same distance in only 6 minut
solmaris [256]

Answer:

6.11km/hr

Step-by-step explanation:

Let the speed that Kelli swims be represented by Y

Speed of the river = 5km/hr

Distance = Speed × Time

Kelli swam upstream for some distance in one hour

Swimming upstream takes a negative sign, hence:

1 hour ×( Y - 5) = Distance

Distance = Y - 5

She then swam downstream the same river for the same distance in only 6 minutes

Downstream takes a positive sign

Converting 6 minutes to hour =

60 minutes = 1 hour

6 minutes =

Cross Multiply

6/60 = 1/10 hour

Hence, Distance =

1/10 × (Y + 5)

= Y/10 + 1/2

Equating both equations we have:

Y - 5 = Y/10 + 1/2

Collect like terms

Y - Y/10 = 5 + 1/2

9Y/10 = 5 1/2

9Y/ 10 = 11/2

Cross Multiply

9Y × 2 = 10 × 11

18Y = 110

Y = 110/18

Y = 6.1111111111 km/hr

Therefore, Kelli's can swim as fast as 6.11km/hr still in the water.

4 0
3 years ago
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