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3 years ago

A Population of 50 Fruit Flies is Increased At A Rate of 6% per Day

Mathematics

1 answer:

andriy [413]3 years ago

5 0

Answer:

Per how many days?

Step-by-step explanation:

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In a metal fabrication process, metal rods are produced to a specified target length of 15 feet. Suppose that the lengths are n

Leto [7]

Answer:

95% Confidence interval: (14.4537 ,15.1463)

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 15 feet

Sample mean, $\bar{x}$ = 14.8 feet

Sample size, n = 16

Alpha, α = 0.05

Sample standard deviation, σ = 0.65 feet

Degree of freedom = n - 1 = 15

95% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 15 and}~\alpha_{0.05} = \pm 2.1314$

$14.8 \pm 2.1314(\dfrac{0.65}{\sqrt{16}} ) \\\\= 14.8 \pm 0.3463 = (14.4537 ,15.1463)$

is the required confidence interval for the true mean length of rods.

3 0

3 years ago

Read 2 more answers

Use an inequality symbol(<>_= to compare _26_54

Rina8888 [55]

26 < 54
26 is less than 54

4 0

3 years ago

Read 2 more answers

Suppose that a group of 10 students are to be assigned to four dorm rooms: a single, a double, a triple and a quad. How many way

Nesterboy [21]

Answer:

Total number of ways to select all sstudent to make assignment = 10×36×70×1 = 25200

Step-by-step explanation:

We have given number of students = 10

These 10 students are assigned to four dorm rooms a single, a double, a triple and a quad

So number of ways to assigned single student $=^{10}C_1=\frac{10!}{1!9!}=10\\$

Now left student = 10 -1 = 9

So number of ways to assigned double student $=^{9}C_2=\frac{9!}{2!7!}=36$

Now left student = 9-2 = 7

So number of ways to assigned triple student $=^{7}C_3=\frac{7!}{2!4!}=$ =70

Now left student = 7 - 3 = 4

So number of ways to assigned quad student $=^{4}C_4=\frac{4!}{4!0!}=1$

So total number of ways to select all sstudent to make assignment = 10×36×70×1 = 25200

6 0

3 years ago

One side of a rectangle is 20 cm larger than the other side. If you make the smaller side two times larger and the larger side t

grandymaker [24]

Let L be the length
Let w be the width
Let p be the perimeter
L+w+L+w=p
L=w+20
3L+2w+3L+2w=240
Sub the first equation in for L in the second equation and solve for w
3(w+20)+2w+3(w+20)+2w=240
3w+60+2w+3w+60+2w=240
10w+120=240
10w=240-120
10w=120
W=120/10
W=12
Sub w into the first equation and solve for L
L=w+20
L=12+20
L=32
Hope this helps!

3 0

3 years ago

Can anyone please help me

inn [45]

9514 1404 393

Answer:

(a) f(3) < g(3)

Step-by-step explanation:

Put the number where the variable is and do the arithmetic.

f(3) = 4·3³ = 4·27 = 108

g(3) = 3·4³ = 3·64 = 192

108 < 192

f(3) < g(3)

8 0

2 years ago