Answer:
the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166
Step-by-step explanation:
The summary of the given statistical data set are:
Sample Mean = 186
Standard deviation = 29
Maximum capacity 3,417 pounds or 17 persons.
sample size = 17
population mean =3417
The objective is to determine the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds
In order to do that;
Let assume X to be the random variable that follows the normal distribution;
where;
Mean [tex]\mu[/tex] = 186 × 17 = 3162
Standard deviation = [tex]29* \sqrt{17}[/tex]
Standard deviation = 119.57
[tex]P(X>3417) = P(\dfrac{X - \mu}{\sigma}>\dfrac{X - \mu}{\sigma})[/tex]
[tex]P(X>3417) = P(\dfrac{3417 - \mu}{\sigma}>\dfrac{3417 - 3162}{119.57})[/tex]
[tex]P(X>3417) = P(Z>\dfrac{255}{119.57})[/tex]
[tex]P(X>3417) = P(Z>2.133)[/tex]
[tex]P(X>3417) =1- 0.9834[/tex]
[tex]P(X>3417) =0.0166[/tex]
Therefore; the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166
A 7 cm x 5 cm rectangle sits inside a circle with radius of 6 cm. What is the area of the shaded region? Round your final answer to the nearest hundredth. Please answer fast!
Answer:78.14cm^2
Step-by-step explanation:
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Enter a number. Round your answer to four decimal places.) μ = 24; σ = 4.2 P(x ≥ 30) = ?
Answer:
Step-by-step explanation:
x is a random variable. Since we are assuming that x has a normal distribution, then we would apply the formula,
z = (x - µ)/σ
Where
x = sample mean
µ = population mean
σ = standard deviation
From the information given,
µ = 24
σ = 4.2
The indicated probability to be determined is P(x ≥ 30)
P(x ≥ 30) = 1 - P(x < 30)
For P(x < 30),
z = (30 - 24)/4.2 = 1.43
Looking at the normal distribution table, the probability corresponding to the z score is 0.9236
Therefore,
P(x ≥ 30) = 1 - 0.9236 = 0.0764
circumference of 6cm ? help plz <3 heyyy b a e (bet you won't reply :)
Answer:
If r = 6 cm, the the circumference is c = 2π(6) = 12π cm
HOPE THIS HELPS AND PLS MARK AS BRAINLIEST
THNXX :)
is lmn congruent to opq if so name the postulate
Answer:
Option (A)
Step-by-step explanation:
Given:
LM ≅ OP
MN ≅ PQ
∠M ≅ ∠P
To Prove:
ΔLMN ≅ ΔOQP
Statements Reasons
1). LM ≅ OP 1). Given
2). MN ≅ PQ 2). Given
3). ∠P ≅ ∠M 3). Given
4). ΔLNM ≅ ΔOQP 4). By the SAS postulate of congruence.
[Side - Angle - Side]
Therefore, Option (A) will be the answer.
An equilateral triangular plate with sides 6 m is submerged vertically in water so that the base is even with the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Round your answer to the nearest whole number. Use 9.8 m/s^2 for the acceleration due to gravity. Recall that the weight density of water is 1000 kg/m^3.) rhog^3(3)^1/2 _______ dx = _______ N
Answer:
26,400 N
Step-by-step explanation:
PLEASE CHECK ATTACHMENT FOR COMPLETE SOLUTION
6 out of 9 pairs of your jeans are blue. What percentage of you
jeans are NOT blue?
Answer:
33.333333%
Step-by-step explanation:
If 6/9 (66.66666666%) of the jeans are blue it means that 3/9 of the jeans are not blue. 3/9 as a percentage is 33.333333%
Answer:
33.3%
Step-by-step explanation:
6 ÷ 9 = 66.66666667%
100% - 66.66666667% = 33.3% (Or you can put 33.33333333)
Hope this helped! :)
Sphere A has a diameter of 2 and is dilated by a scale factor of 3 to create sphere B. What is the ratio of the volume of sphere A to sphere B? 2:6 4:36 1:3 1:27
Answer:
1:27 (D)
Step-by-step explanation:
Given:
Sphere A has a diameter of 2
Sphere A is dilated to create sphere B
Scale factor = 3
Volume of a sphere = 4/3 πr³
Radius = r = diameter/2 = 2/2
r = 1
Volume of sphere A = 4/3 ×π(1)³
Volume of sphere A = 4/3 × π
Volume of sphere B = 4/3 πR³
Since the diameter was dilated, the diameter of B = diameter of A × scale factor
diameter of B = 2×3 = 6
Radius of B = R = diameter/2 = 3
Volume of sphere B = 4/3 × π(3)³
Volume of sphere B = (4/3)(27)π
Ratio of the volume of sphere A to volume of sphere B
= [4/3 ×π]: [(4/3)(27)π]
= (4π/3)/[(4π/3)×27] = 1/27
= 1:27
Answer: 1:27
Step-by-step explanation:
The original volume * scale factor cubed = new volume.
The scale factor is 3 and 3^3 is 27, so the ratio is 1:27
Couple more! Running out of time lol!
Answer:
A translation; (x,y) --> (x-4,y-5)
Step-by-step explanation:
This is because the figures are congruent and in the same orientation but just in different locations on the coordinate plane.
A(0,3) --> A'(-4,-2)
So, the rule is (x,y) --> (x-4,y-5)
The quick ratio, Q, is calculated using the formula Q= CA-I-P/ CL, where CA is the value of the company’s current assets, I is inventory, P is prepaid expenses, and CL is current liabilities. Rearrange the formula for current assets
Answer:
CA = Q·CL +I +P
Step-by-step explanation:
Multiply by the denominator, then add the opposite of all terms that are not CA.
[tex]Q=\dfrac{CA-I-P}{CL}\\\\Q\cdot CL=CA-I-P\\\\Q\cdot CL+I+P=CA\\\\\boxed{CA=Q\cdot CL+I+P}[/tex]
Name the triangle with the following characteristics. sides: 5 cm, 6 cm, 7 cm; Angles: 75° and 60°. yeah
Answer:
Step-by-step explanation:
this triangle is regular one
we can't apply the pytahgorian theorem 5²+6²≠7² the angles have different sizes 75≠60≠45the sides have different lengths 5≠6≠7Answer:
obtuse scalene triangle
What is the measure of angle ABC? Please answer quickly!
Answer:
ABC = 88
Step-by-step explanation:
Angle Formed by Two Chords = 1/2( sum of Intercepted Arcs)
ABD = 1/2 ( 131+ 53)
ABD = 1/2 (184)
=92
ABC = 180 -ABD
ABC = 88
When testing for current in a cable with sixsix color-coded wires, the author used a meter to test threethree wires at a time. How many different tests are required for every possible pairing of threethree wires?
Answer:
20 tests
Step-by-step explanation:
There are six different wires, and for each test the author picks three, this means that each test is a combination of three out of six wires (₆C₃) . The number of total combinations possible is given by:
[tex]_6C_3=\frac{6!}{(6-3)!3!}\\_6C_3=\frac{6*5*4}{3*2*1}\\_6C_3=20[/tex]
20 tests are required to verify every possible pairing of three wires.
Find value of z and simplify completely.
Answer:
3√10Given:
A right triangle in which an altitude is drawn from the right angle vertex
To find: value of X
We have leg rule in similarity in right triangle as:
[tex] \frac{leg}{part} = \frac{hypotenuse}{leg} \\ [/tex]
Plugging the given values,
[tex] \frac{z}{3} = \frac{3 + 27}{z} \\ z \times z = 3(3 + 27) \: \: (cross \: multiplication) \\ {z}^{2} = 9 + 81 \\ {z}^{2} = 90 \\ z = \sqrt{90} \\ z = 3 \sqrt{10} [/tex]
Hope this helps...
Good luck on your assignment..
Please answer this correctly
it would be possible because if it's 6 it's even if it 4 it's even if it's 2 it's even so you okay so it is certain and possible babes
Pls help me pick the right answer! Please
A camera shop stocks eight different types of batteries, one of which is type A76. Assume there are at least 30 batteries of each type.
Required:
a. How many ways can a total inventory of 30 batteries be distributed among the eight different types.
b. How many ways can a total inventory of 30 batteries be distributed among the eight different types if the inventory must include at least four A76 batteries?
c. How many ways can a total inventory of 30 batteries be distributed among the eight different types if the inventory includes at most three A7b batteries?
Answer:
a. 10295472 ways
b. 4272048 ways
c. 6023424 ways
Step-by-step explanation:
Given that:
Camera shop stocks ----- 8 different types of batteries
one of which is ---- A76
Assume that there are ------ at least 30 batteries of each type.
a.
How many ways can a total inventory of 30 batteries be distributed among the eight different types.
The number of ways a total inventory of 30 batteries be distributed is :
[tex]= \left \{ {{30+8-1} \atop {30}} \right. \}[/tex]
[tex]= \left \{ {{37} \atop {30}} \right. \}[/tex]
[tex]=\dfrac{37!}{30! *7!}[/tex]
[tex]= \dfrac{37*36*35*34*33*32*31*30!}{30!*7*6*5*4*3*2*1}[/tex]
[tex]= \dfrac{37*36*35*34*33*32*31}{7*6*5*4*3*2*1}[/tex]
= 10295472 ways
b.
How many ways can a total inventory of 30 batteries be distributed among the eight different types if the inventory must include at least four A76 batteries?
If we must include 4 A76 batteries; then the number of ways a total inventory of 30 batteries can be distributed among eight different types of batteries will be:
30 - 4 = 26 batteries
Now;
[tex]= \left \{ {{26+8-1} \atop {26}} \right. \}[/tex]
[tex]= \left \{ {{33} \atop {26}} \right. \}[/tex]
[tex]=\dfrac{33!}{26! \ \ 7!}[/tex]
[tex]=\dfrac{33*32*31*30*29*28*27*26!}{26! \ * \ 7*6*5*4*3*2*1}[/tex]
[tex]=\dfrac{33*32*31*30*29*28*27}{ \ 7*6*5*4*3*2*1}[/tex]
= 4272048 ways
c. If we must include at most three A7b batteries. the number of ways that a total inventory of 30 batteries can be distributed among eight different types of inventory is:
[tex]= \sum \limits ^3 _{x=0} \left \{ {{(30-x)+7-1} \atop {30-x}} \right. \} \\ \\ \\ = \sum \limits ^3 _{x=0} \left \{ {{(30-0)+7-1} \atop {30-0}} \right. \} = (^{36}_{30})+(^{35}_{29})+ (^{34}_{28})+ (^{33}_{27})[/tex]
[tex]= \dfrac{36!}{30! * 6!} + \dfrac{35!}{29! * 6!} + \dfrac{34!}{28! * 6!} + \dfrac{33!}{27! * 6!}[/tex]
= 1947792 + 1623160 + 1344904 + 1107568
= 6023424 ways
If the area to the left of x in a normal distribution is 0.187, what is the area to the right of x? (Enter an exact number as an integer, fraction, or decimal.)
Answer:
Area to the right is 0.813
Step-by-step explanation:
Since the area under the full normal distribution should render "1" (one), then the area to the right of x should be:
1 - area to the left of x = 1 - 0.187 = 0.813
The area to the right of x in this normal distribution will be 0.813.
What is a normal distribution?The majority of the observations are centered around the middle peak of the normal distribution, which is a continuous probability distribution that is symmetrical around its mean.
The probabilities for values that are farther from the mean taper off equally in both directions. Extreme values in the distribution's two tails are likewise rare.
Not all symmetrical distributions are normal, even though the normal distribution is symmetrical.
We know that the total area described in a normal distribution is 1.
Given, that the area to the left of x in a normal distribution is 0.187.
So, the area to the right of x will be,
1 - 0.187.
= 0.813.
learn more about normal distribution here :
https://brainly.com/question/15103234
#SPJ2
Factor by grouping 5x^3+6x^2+25x+30
Answer:
(5x + 6) (x² + 5)
Step-by-step explanation:
5x³ + 6x² + 25x + 30
= x² (5x + 6) + 25x + 30 -- Group 5x³ and 6x²
= x² (5x + 6) + 5 (5x + 6) -- Group 25x and 30
= (5x + 6) (x² + 5) -- Both terms have a common factor of 5x + 6
You buy a 33-pound bag of flour for $9 or you can buy a 1- pound bag for $0.39. Compare the per pound cost for the large and small bag. How much is the pounds per dollar
Answer:
see below
Step-by-step explanation:
9 dollars / 33 lbs = .272727 dollars per lb
.39 / 1 lbs = .39 per lb
The large bag is less expensive
Step-by-step explanation:
33lb bag = $9.00
1lb = $0.39
$9.00 bag per pound is $.0.27 per pound
If I bought 33, 1 pound bags it would cost $12.87
cheaper by $ 3.87 to buy the 9lb bag.
Each of 100 students in the Allen School can only take 1 CSE class each, between the four classes CSE 311, CSE 312, CSE 331, and CSE 332. Each student (independently of others) takes CSE 311 with probability 0.3, CSE 312 with probability 0.4, CSE 331 with probability 0.1, and CSE 332 with probability 0.2. What is the probability that exactly 31 sign up for CSE 311, 39 sign up for CSE 312, 7 sign up for CSE 331, and 23 sign up for CSE 332
Answer:
[tex]P(a=31,b=39,c=7,d=23) = 0.000668[/tex]
Step-by-step explanation:
Sample space, n = 100
Let the number of students signed up for CSE 311 = a
Let the number of students signed up for CSE 312 = b
Let the number of students signed up for CSE 331 = c
Let the number of students signed up for CSE 332 = d
Probability of taking CSE 311, [tex]P_a[/tex] = 0.3
Probability of taking CSE 312, [tex]P_b[/tex] = 0.4
Probability of taking CSE 331, [tex]P_c[/tex] = 0.1
Probability of taking CSE 332, [tex]P_d[/tex] = 0.2
[tex]P(a,b,c,d) = \frac{n!}{a! b! c! d!} p_a^{a} p_b^{b} p_c^{c} p_d^{d} \\P(a=31,b=39,c=7,d=23) = \frac{100!}{31! 39! 7! 23!} * 0.3^{31} * 0.4^{39} * 0.1^{7} 0.2^{23}\\P(a=31,b=39,c=7,d=23) = \frac{4.58*10^{111}}{2.13*10^{56}* 5040 }* (1.57*10^{-55})\\P(a=31,b=39,c=7,d=23) = 0.000668[/tex]
PLEASE HELPP! f(x)= -3x + 3
Which of the graphs represent the inverse of the function F??
Answer:
Answer is Y
Step-by-step explanation:
Please answer this correctly
Answer:
0
Step-by-step explanation:
The probability of picking a even number is 1/3
If u don’t replace it then the probability of picking an even number is 0/3
Multiply and u get 0/9 or 0
Hope this helps
Answer:
0
Step-by-step explanation:
The number 2 is even.
The probability of picking an even number is 1/3.
You don't put the first card back.
1 and 3 are odd.
1/3 × 0 = 0
A pet store has 32 more dogs than cats. If there are a total of 144 dogs and cats altogether, then how many cats does the pet store have?
PLZ ANSWER ASAP THANKS
Answer:
56 cats
88 dogs
Step-by-step explanation:
x=cats
x+32=dogs
x+x+32=144
2x=144-32
2x=112
x=112/2
x=56 (56 cats)
x+32=88 (88 dogs)
Answer:
56
Step-by-step explanation:
In order to solve this question you need to subtract 32 by 144 then divide that answer by 2.
So we know that they're 32 more dogs then cats and there is 144 dogs and cats all together.
[tex]144 - 32 = 112[/tex]
Divide it in half
[tex]112 \div2=56[/tex]
[tex]56+32 = 88[/tex]
[tex]= 56[/tex]
Therefore they're 56 cats in the pet store.
Hope this helps.
Simplify The square root of 5 (6-4 the square root of 3)
Answer:
7.75
Step-by-step explanation:
6-4=2
2 times the square root of 3=3.46410161514
square root of 5 times 3.46410161514=7.74596669242
to 2dp=7.75
Progress
Question ID: 470099
One student can paint a wall in 12 minutes. Another student can paint the same wall in 24 minutes. Working together, how long will it
take for them to paint the wall?
Answer:
8 minStep-by-step explanation:
Try this:
1 wall 1 288
------------------------ = --------------- = --------------- min = 8 min
1 wall 1 wall 24 + 12 36
(---------) + (---------) ------------
12 min 24 min 288
Reflected across the x-axis ? How is look in the figure
Answer:
Well a shape reflecting across the x axis will look like some thing in the image below.
Look at both images.
Simplify e^ln4
A. 1/4
B. 4
C. 1n4
D. E^4
Answer:
The answer is option B.
4Step-by-step explanation:
Using the expression
[tex] {e}^{ ln(x) } = x[/tex]
[tex] {e}^{ ln(4) } = 4[/tex]
Hope this helps you
Which of the following theorems verifies that WVU= RST
Answer:
C. HL
Step-by-step explanation:
The Hypotenuse-Leg Theorem is the only viable way to determine congruency between 2 right triangles.
Tom takes a cancer test and the test is advertised as being 99% accurate: if you have cancer you will test positive 99% of the time, and if you don't have cancer, you will test negative 99% of the time. If 1% of all people have cancer and Tom tests positive, what is the prob that Tom has the disease
Answer:
99% chance tommy has it
Step-by-step explanation:
cuz do da math
"a. How many study subjects were cases? b. How many study subjects were controls? c. What was the ratio of controls to cases?"
Answer:
The description is provided following.
Step-by-step explanation:
The given question is incomplete. The complete question will be:
Brain tumors No Brain tumors
Cell Phones 63 185
No Cell Phones 96 292
The further explanation is given below.
a...
Subjects with these symptoms/diseases are recognized as "cases." Consequently, the majority of the instances would be as follows:
⇒ [tex]63+96[/tex]
⇒ [tex]159[/tex]
b...
Subjects who might not have the disorder or infection are classified as "controls." Therefore, the amount of controls is as follows:
⇒ [tex]185+292[/tex]
⇒ [tex]477[/tex]
c...
The proportion of control and monitoring of instances:
⇒ [tex]\frac{478}{159}[/tex]
⇒ [tex]3.006[/tex]