What is the probability that the two pointers will land on an odd number and letter C? Give the probability as a percent

Answer:

speak

Step-by-step explanation:

Answer:

b=90

c=30a=60

this kdkddkdkdk

Answer:

[tex] \sqrt{2} [/tex]

Hope it helps

pls give brainliest

Answer:

i think the answer should be 4y(m+3)/a^2

Step-by-step explanation:

Answer:

37.5x

Step-by-step explanation:

Answer:

(-6,2)

Step-by-step explanation:

when it asks for the scale factor all you do is take your coordinates and multiply it by whatever the question gives you. Like in this question you would multiply -3 and 1 by 2 and whatever the number comes out to be then thats your answers.

Sorry if the explanation isnt very good!

Answer:

Distance in ft = 597.57 feet

Step-by-step explanation:

As per the

First equation of motion

84 mph = 37.5514 m/s

45 mph = 20.1168 m/s

v-u = a* t

Substituting the given values, we get -

37.5514 m/s - 20.1168 m/s = a * 6.3

a = 2.76 m/s^2

Now as per Newton's third law of motion

v^2-u^2 = 2 *a*S

Substituting the given values, we get -

37.5514^2 - 20.1168^2 = 2 * 2.76 m/s^2 * S

S = 182.142 Meters

Distance in ft = 597.57 feet

Answer:

347,392

Step-by-step explanation:

yes

Answer:

answer B

Step-by-step explanation:

Hope this helps

Answer:

For a data set of N elements:

{x₁, x₂, ..., xₙ}

The median is the middle value (or the average of the two middle values)

The average or mean is:

[tex]m = \frac{x_1 + ... + x_n}{N}[/tex]

The mode is the value that is more repeated.

The range is the difference between the largest value and the smallest value.

Now, let's try to construct two sets A and B that meet the required conditions:

Let's suppose two sets of 5 values each:

A = {2, 2, 3, 4, 5}

The median of A is 3.

The range of A is = 5 - 2 = 3

The average of A is: (2 + 2 +3 +4 +5)/5 = 16/5

And the value "2" appears two times, so the mode is 2.

Now, let's try to find a data set B such that:

The mode is smaller than 2  (so for example, we can have a "1" that repeats two times)

The median is smaller than 3 (let's have a median equal to 2)

The average is larger than 16/5

The range is larger than 3

This is ratter simple:

B = {1, 1, 2, 6, 9}

The median of B is 2.

The mode of B is 1

The range of B is: 9 - 1 = 8

The average of B is: (1 + 1 + 2 + 6 + 9)/5 = 19/5

So, we just found an example for two data sets A and B such that A has a greater median and mode, while B has a greater average and range.

So the answer is yes, it is possible.

Answer:

>

Step-by-step explanation:

Answer:

Because there are 9 sour straws in total and 3 blue sour straws, the probability of pulling out a blue sour straw would be 3/9 or 1/3. Because we aren't replacing it, that means that there's one less straw in the box so there should only be 8 straws left. Because there's 4 pink sour straws, the probability of pulling out a pink sour straw is 4/8 or 1/2.

Answer:

Volume of iron used is 58982cm

Step-by-step explanation:

For hollow cylinder,

\bold{Height=9cm}Height=9cm

\bold{Internal \ radius=21cm}Internal radius=21cm

\bold{Externa \ radius=28cm}Externa radius=28cm

volume \: of \: cylinder = \pi \times {r}^{2} \times h \:volumeofcylinder=π×r

2

×h

\bold{Volume \ of \ cylinder \ with \ hollow \ part}Volume of cylinder with hollow part

\bold{=\frac{22}{7}×28^2×9}=

7

22

×28

2

×9

\bold{=22×4×28×9}=22×4×28×9

\bold{=71456cm^3}=71456cm

3

\bold{Volume \ of \ hollow \ part}Volume of hollow part

\bold{=\frac{22}{7}×21^2×9}=

7

22

×21

2

×9

\bold{=22×3×21×9}=22×3×21×9

\bold{=12474cm^3}=12474cm

3

\bold{Volume \ of \ iron \ used=}Volume of iron used=

\bold{Volume \ of \ cylinder \ with \ hollow \ part}Volume of cylinder with hollow part

\bold{- Volume \ of \ hollow \ part}−Volume of hollow part

{=71456-12474}=71456−12474

{=58982cm^3}=58982cm

3

\bold\red{Volume \ of \ iron \ used \ is \ 58982cm^3}Volume of iron used is 58982cm

3

Answer:

The man has made a profit, for with the same money he bought more British Pounds than he could have bought a week later.

Step-by-step explanation:

Given that a man converted Rs. 150000 into pound sterling, and after one week Nepalese rupees devalued by 5%, to find if you have obtained a profit or a loss, the following calculation must be performed:

150,000 / 77.02 = X

1,947.55 = X

150,000 / (77.02 x 1.05) = X

150,000 / 80.87 = X

1,854.82 = X

Therefore, the man has made a profit, for with the same money he bought more British Pounds than he could have bought a week later.

Answer:

0.1764

Step-by-step explanation:

This problem meets requirement that enables us to use the binomial probability relation :

Probability of success, p = 17% = 0.17

number of trials = 9

(1 - p) = 1 - 0.17 = 0.83

The binomial probability relation :

P(x = x) = nCx * p^x * (1 - p)^(n-x)

P(x = 3) = 9C3 * 0.17^3 * 0.83^6

P(x = 3) = 84 * 0.001606258054361897

P(x = 3) = 0.134925676566399348

P(x = 3) = 0.13492

P(x = 4) = 9C4 * 0.17^4 * 0.83^5

P(x = 4) = 126 * 0.000328992613544003

P(x = 4) = 0.041453069306544378

P(x = 4) = 0.04145

P(x = 3 or x = 4) = P(x = 3) + p(x = 4)

= 0.13492 + 0.04145

= 0.17637

= 0.1764

yo same, im super lost in life too, my parents want one thing, and i want another, like cmon, let me live give me a break....

Step-by-step explanation:

Answer:

11 [tex]\frac{25}{64}[/tex]

Step-by-step explanation:

volume is length*width*height

and we have a cube( whenever you have a cube, its length, width, and height are all equal) so 2.25*2.25*2.25=11 25/64

Answer:

x=[tex]\frac{4}{5}[/tex]

y=[tex]\frac{23}{5}[/tex]

Step-by-step explanation:

Look at the pictures!

Answer:

Solution given:

(x1,y1)=(-3,2)

(x2,y2)=(2,2)

length[d]=?

we have

d=[tex] \sqrt{(x2-x1)²+(y2-y1)²} [/tex]

=[tex] \sqrt{(2+3)²+(2-2)²} [/tex]

=[tex] \sqrt{5²} [/tex]

=5units

D 5 units is a right answer.

Answer B

Distribute the 2

Answer:

C, D and F

Step-by-step explanation:

Procedure:

9x+5=5x+21

4x+5=21 (subtract 5x)

4x=16 (subtract 5)

x=4 (divided by 4)

Answer:

a) probability that any individual student has a flu shot is  0.30

b) The probability that more than 30 students have had flu shots is 0.1038

c) the probability that 20 or fewer students have had shots is 0.1314

Step-by-step explanation:

Given the data in the question;

Number of students n = 84

student health clinic reports that only 30% of students got a flu shot this year;

p = 30% = 0.30

now,

pn = 0.30 × 84 = 25.2 ( > 10 )

qn = (1-p)n = (1-0.30)84 = 0.70 × 84 = 58.8  ( > 10 )

Criterion for approximation of the binomial distribution is satisfied since both pn and qn are greater than 10.

hence, Normal distribution will have;

mean μ = np = 0.30 × 84 = 25.2

standard deviation σ = √( npq )

= √( 84 × 0.30 × 0.70 )

= √17.64

standard deviation σ = 4.2

a) the probability that any individual student has a flu shot

student health clinic reports that only 30% of students got a flu shot this year

p = 30% = 0.30

Therefore, probability that any individual student has a flu shot is  0.30

b) The probability that more than 30 students have had flu shots

P(X > 30)

we use real time limits to determine probability from binomial distribution,

hence upper limit will be 30.5

Z = (X - np) / √( npq )

Z = ( 30.5 - 25.2 ) / 4.2

Z = 5.3 / 4.2

Z = 1.26

from normal distribution table

[tex]Z_{1.26[/tex] = 0.1038

Therefore, The probability that more than 30 students have had flu shots is 0.1038

c) the probability that 20 or fewer students have had shots;

P( X ≤ 20 )

upper limit will be 20.5

Z = (X - np) / √( npq )

Z = ( 20.5 - 25.2 ) / 4.2

Z = -4.7 / 4.2

Z = -1.12

from normal distribution table

[tex]Z_{-1.12[/tex] = 0.1314

Therefore, the probability that 20 or fewer students have had shots is 0.1314

Answer:

160

Step-by-step explanation:

20 x 2 = 40

40 x 2 = 80

80 x 2 = 160

It stops at 3 years.

Answer:

120

Step-by-step explanation:

20+40+60 = 120

If we double it, we have to keep adding 20 each year and stop at 3 so the answer is 120

Answer: $2

Step-by-step explanation:

Answer:

ΔSTU ≅ ΔBDC

Step-by-step explanation:

In ΔSTU and ΔBDC,

∠S ≅ ∠B [Given]

∠T ≅ ∠D [Given]

SU ≅ BC [Given]

Since, two corresponding angles and non included side of the angles are equal in measure.

Therefore, ΔSTU ≅ ΔBDC [By AAS property of congruence]