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3 years ago

PLS HURRY THIS IS REALLY HARD!(FOR ME)

Mathematics

1 answer:

babunello [35]3 years ago

8 0

Answer:

She needs about 398 more dollars

Step-by-step explanation: The price of the whiteboard is $989. Add the money from what she collected and the fundraising. 485+106 then you take that and subtract it from 989.

Hope this helped!

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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= X[bar] - μ ~ 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

What’s the answer to x

grigory [225]

Answer:

x = 55

Step-by-step explanation:

a triangle's angles must equal when added up to 180.

76 + 49 + x = 180

combine terms

125 + x = 180

x = 55

7 0

3 years ago

If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.

Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have $S_1=3\times2^{1-1}+2(-1)^1=3-2=1$ , which agrees with the initial value S₁ = 1.

• Assume the closed form is correct for all n up to n = k. In particular, we assume

$S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}$

and

$S_k=3\times2^{k-1}+2(-1)^k$

We want to then use this assumption to show the closed form is correct for n = k + 1, or

$S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}$

From the given recurrence, we know

$S_{k+1}=S_k+2S_{k-1}$

so that

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)$

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}$

$S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)$

$S_{k+1}=3\times2^k-2(-1)^k$

$S_{k+1}=3\times2^k+2(-1)(-1)^k$

$\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}$

which is what we needed. QED

6 0

3 years ago

Find the modulus of the complex number 6-2i

Papessa [141]

Answer:

The modulus of the complex number 6-2i is:

$|z|\:=2\sqrt{10}$

Step-by-step explanation:

Given the number

$6-2i$

We know that

$z = x + iy$

where x and y are real and $\sqrt{-1}=i$

The modulus or absolute value of z is:

$|z|\:=\sqrt{x^2+y^2}$

Therefore, the modulus of $6-2i$ will be:

$z=6-2i$

$z=6+(-2)i$

$|z|\:=\sqrt{x^2+y^2}$

$|z|\:=\sqrt{6^2+\left(-2\right)^2}$

$=\sqrt{6^2+2^2}$

$=\sqrt{36+4}$

$=\sqrt{40}$

$=\sqrt{2^2}\sqrt{2\cdot \:5}$

$=2\sqrt{2\cdot \:5}$

$=2\sqrt{10}$

Therefore, the modulus of the complex number 6-2i is:

$|z|\:=2\sqrt{10}$

3 0

2 years ago

Solve for x: 14-14x^1/3 =2

zmey [24]

Answer: 216/343 or 0.6297

Step-by-step explanation:

$14-14x^{1/3} = 2\\14 -2 = 14x^{1/3}\\12 = 14x^{1/3}\\\frac{12}{14} =x^{1/3}\\\frac{6}{7} =x^{1/3}\\(\frac{6}{7})^3 =(x^{1/3})^3\\\frac{6^3}{7^3} =x\\\\\frac{216}{343} = 0.629737609 = x$

7 0

3 years ago

PLS HURRY THIS IS REALLY HARD!(FOR ME)​

PLS HURRY THIS IS REALLY HARD!(FOR ME)