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3 years ago

Can u help with these two

Mathematics

2 answers:

Artemon [7]3 years ago

8 0

The volume of a rectangular prism can be given by the form
$V= b*h*w$

you know Volume, base, and height but you dont know width

so divide b and h to both sides
$\frac{V}{b*h} = w$
and you cna solve for width

now for 6.
dividing by a number is the same as multiplying by its reciprocal

$n \div 6 = n * \frac{1}{6}$

NemiM [27]3 years ago

7 0

1/8 divide 6/1= 1/48
now needs to change to multiplying. So 6/1 goes to reciprocal 1/6.

1/8 X 1/6 =1/48

Hope this helps!!

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Shobo's mother's present age is six times Shobo's present age. Shobo's age five

il63 [147K]

Answer:

Step-by-step explanation:

Shobo's age = x years

His mother's age = 6 times of

Shobo's age

= 6x years

Five years from now

________________

Shobo's age = ( x + 5 ) years

According to the problem given,

x + 5 = one by third of his mother's

present age

x + 5 = 6x / 3

x + 5 = 2x

5 = 2x - x

5 = x

Therefore ,

Shobo's present age = x = 5 years

His mother's age = 6x

= 6 × 5

= 30 years

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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