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Sophie [7]
3 years ago
11

Text me the answer please

Mathematics
2 answers:
svetoff [14.1K]3 years ago
8 0
P(4)= 1/8

This is because there are 8 equal parts and 4 is just one of those 8 parts.
Arte-miy333 [17]3 years ago
6 0
Probablility=desired outcomes/total possible outcomes

so

we want to hit the 4

there are 1 of those numbers on the spinner
and 8 numbers total

1 desired outcome
8 total possible outcomes

1/8 is the probablitiy
?=8
You might be interested in
A mass of 1 g is set in motion from its equilibrium position with an initial velocity of 6in/sec, with no damping and a spring c
yan [13]

a) y(t)=0.0016 sin(94.9t) [m]

b) 0.033 s

c) -0.152 m/s

Step-by-step explanation:

a)

The force acting on the mass-spring system is (restoring force)

F=-ky

where

k = 9 is the spring constant

y is the displacement

Also, from Newton's second law of motion, we know that

F=my''

where

m = 1 g = 0.001 kg is the mass

y'' is the acceleration

Combining the two equations,

my''=-ky

This is a second order differential equation; the solution for y(t) is

y(t)=A sin(\omega t-\phi)

where

A is the amplitude of motion

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{9}{0.001}}=94.9 rad/s is the angular frequency

The spring starts its motion from its equilibrium position, this means that y=0 when t=0; therefore, the phase shift must be

\phi=0

So the displacement is

y(t)=A sin(\omega t)

The velocity of the spring is equal to the derivative of the displacement:

v(t)=y'(t)=\omega A cos(\omega t)

We know that at t = 0, the initial velocity is 6 in/s; since 1 in = 2.54 cm = 0.0254 m,

v_0=6(0.0254)=0.152 m/s

And since at t = 0, cos(\omega t)=1

Then we have:

v_0=\omega A

From which we find the amplitude:

A=\frac{v_0}{\omega}=\frac{0.152}{94.9}=0.0016 m

So the solution for the displacement is

y(t)=0.0016 sin(94.9t) [m]

b)

Here we want to find the time t at which the mass returns to equilibrium, so the time t at which

y=0

This means that

sin(\omega t)=0

We know already that the first time at which this occurs is

t = 0

Which is the beginning of the motion.

The next occurence of y = 0 is instead when

\omega t = \pi

which means:

t=\frac{\pi}{\omega}=\frac{\pi}{94.9}=0.033 s

c)

As said in part a), the velocity of the mass-spring system at time t is given by the derivative of the displacement, so

v(t)=\omega A cos(\omega t)

where we have

\omega=94.9 rad/s is the angular frequency

A=0.0016 m is the amplitude of motion

t is the time

Here we want to find the velocity of the mass when the time is that calculated in part b):

t = 0.033 s

Substituting into the equation, we find:

v(0.033)=(94.9)(0.0016)cos(94.9\cdot 0.033)=-0.152 m/s

4 0
3 years ago
Can anyone help me?
ArbitrLikvidat [17]

Answer:

The factored form of 4<em>m</em>³ – 28<em>m</em>² – 120<em>m</em> is 4<em>m</em>(<em>m</em> – 10)(<em>m</em> + 3). The zeroes of the function would be <em>m</em> = 0, <em>m</em> = –3, and <em>m</em> = 10.

Step-by-step explanation:

I'll give this a shot.

4<em>m</em>³ – 28<em>m</em>² – 120<em>m</em> = 0 — The original expression

4<em>m</em>³ – 28<em>m</em>² – 120<em>m</em> — Did that 0 have any purpose? I just deleted it.

4(<em>m</em>³ – 7<em>m</em>² – 30<em>m</em>) — There's a common factor in here, 4. Let's pull that aside.

4m(<em>m²</em> – 7<em>m</em> – 30) — Actually, there's <em>two</em> common factors. The second one is <em>m</em>! Let's pull <em>that</em> out too!

To factor an expression, you have to break apart the middle term, so to speak. That's only possible if you can find two numbers whose product equals that of the outside terms and whose sum equals the middle term. Here, I'm just dealing with numbers and putting that variable aside.

–30 = 10 × –3

–30 = –10 × 3

–30 = –2 × 15

–30 = 2 × –15 — To solve for any potential factors, let's find all the numbers integers that multiply to –30

Now let's see which one adds up to –7!

15 – 2 = 13 — it's not this one

2 – 15 = –13 — nor this one

10 – 3 = 7 — we're pretty close! Let's switch that negative

3 – 10 = –7 — here we go! Here's our numbers!

4<em>m</em>[(<em>m</em>² – 10<em>m</em>) + (3<em>m</em> – 30)] — now we break apart the middle term. This is <em>all</em> multiplied by 4<em>m</em>, so that still encases everything with brackets.

4<em>m</em>[<em>m</em>(<em>m</em> – 10) + 3(<em>m</em> – 10)] — Factoring the two expressions

4<em>m</em>(<em>m</em> + 3)(<em>m</em> – 10) — simplifying to find our answer! Ta-da!

8 0
3 years ago
In a history class, the girl to boy ratio is 5 to 3. If there are 80 students total, how many boys are there?​
Brilliant_brown [7]
80x3/8=10 common factor. 10x5. And 10x3


50/30

30 boys

50 girls
8 0
2 years ago
HELP ASAP!
kolezko [41]

Because vertical angles always equal each other. linear pairs always equals 180 degrees always. hope this helps :)

8 0
3 years ago
A quality analyst of a tennis racquet manufacturing plant investigates if the length of a junior's tennis racquet conforms to th
Burka [1]

Answer:

Confidence interval : 21.506 to 24.493

Step-by-step explanation:

A quality analyst selects twenty racquets and obtains the following lengths:

21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

So, sample size = n =20

Now we are supposed to find Construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

Since n < 30

So we will use t-distribution

Confidence level = 99.9%

Significance level = α = 0.001

Now calculate the sample mean

X=21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

Sample mean = \bar{x}=\frac{\sum x}{n}

Sample mean = \bar{x}=\frac{21+25+23+22+24+21+25+21+23+ 26+ 21+24+22+ 24+23+21+ 21+ 26+23+ 24}{20}

Sample mean = \bar{x}=23

Sample standard deviation = \sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}

Sample standard deviation = \sqrt{\frac{(21-23)^2+(25-23)^2+(23-23)^2+(22-23)^2+(24-23)^2+(21-23)^2+(25-23)^2+(21-23)^2+(23-23)^2+(26-23)^2+(21-23)^2+(24-23)^2+(22-23)^2+(24-23)^2+(23-23)^2+(21-23)^2+(21-23)^2+(26-23)^2+(23-23)^2+(24-23)^2}{20-1}}

Sample standard deviation= s = 1.72

Degree of freedom = n-1 = 20-1 -19

Critical value of t using the t-distribution table t_{\frac{\alpha}{2} = 3.883

Formula of confidence interval : \bar{x} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}

Substitute the values in the formula

Confidence interval : 23 \pm 1.73 \times \frac{1.72}{\sqrt{20}}

Confidence interval : 23 -3.883 \times \frac{1.72}{\sqrt{20}} to 23 + 3.883 \times \frac{1.72}{\sqrt{20}}

Confidence interval : 21.506 to 24.493

Hence Confidence interval : 21.506 to 24.493

3 0
3 years ago
Read 2 more answers
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