9514 1404 393
Answer:
C. 837
Step-by-step explanation:
The remaining angle is ...
C = 180° -A -B = 180°-62° -67° = 51°
The law of sines tells us that the length AC is ...
AC/sin(B) = AB/sin(C)
AC = AB·sin(B)/sin(C) = 40·sin(67°)/sin(51°)
Using the area formula given, we now have ...
area = 1/2(AB)(AC)sin(A)
= (1/2)(40)(40·sin(67°)sin(62°)/sin(51°) ≈ 836.7
The area of the triangle is about 837 square units.
Here's the result of this question
The point (-2,7) has undergone the following transformations:
1. Translated 1 unit up and 4 units left
Then
2. Reflected about the c-axis
Then
3. Rotated 90° anticlockwise about the origin
A) Its final coordinates are (3,-1)
B) Its final coordinates are (8,-6)
C) Its final coordinates are (-8,6)
D) Its final coordinates are (-3,1)
Answer:
B) Its final coordinates are (8,-6)
Step-by-step explanation:
1. Translated 1 unit up and 4 units left
(-2,7) becomes (-6, 8)
2. Reflected about the x-axis
(-6,8) becomes (-6, -8)
3. Rotated 90° anticlockwise about the origin
(-6, -8) becomes (8, -6) because when rotating 90 degrees anticlockwise about the origin, point A (x,y) becomes point A' (-y,x). In other words, switch the x and y and make y negative.
The principle
P=6000 A=6810 T=3 years
Answer:
incomplete question
Step-by-step explanation:
that is what is wrong with your question
Answer:
r = 4.3%
Step-by-step explanation:
6810= 6000(x)^3
6810/6000= (x)^3
x = 1.043114431
r = 043114431
a recent survey shows that 66% of college students have a cat and 37% have a HBO subscription. Assuming these two events are independent, what is the probability that a randomly selected student has neither a cat nor HBO
Answer:
[tex]P(C'\ and\ H') =0. 2178[/tex]
Step-by-step explanation:
Let
[tex]C \to[/tex] Student with cat
[tex]H \to[/tex] Student has HBO sub
[tex]P(C) = 66\% \\ P(H) = 37\%[/tex]
Required
[tex]P(C'\ and\ H')[/tex]
This is calculated as:
[tex]P(C'\ and\ H') = P(C') * P(H')[/tex]
Using complement rules, we have:
[tex]P(C'\ and\ H') = [1 - P(C)] * [1 - P(H)][/tex]
So, we have:
[tex]P(C'\ and\ H') = [1 - 66\%] * [1 - 37\%][/tex]
[tex]P(C'\ and\ H') = [33\%] * [66\%][/tex]
[tex]P(C'\ and\ H') =0. 2178[/tex]
A trader sold 90 oranges at 3 for GHC 0.75.
How much did she get from selling all the
oranges?
Answer:
GHC22.5
Step-by-step explanation:
90/3=30
30=0.75
30×0.75
=22.5
Calls to a customer service center last on average 2.3 minutes with a standard deviation of 2 minutes. An operator in the call center is required to answer 76 calls each day. Assume the call times are independent.
What is the expected total amount of time in minutes the operator will spend on the calls each day?
What is the standard deviation of the total amount of time in minutes the operator will spend on the calls each day? Give your answer to four decimal places.
What is the approximate probability that the total time spent on the calls will be less than 166 minutes? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question.
What is the value c such that the approximate probability that the total time spent on the calls each day is less than c is 0.95? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question.
Answer:
The expected total amount of time in minutes the operator will spend on the calls each day is of 174.8 minutes.
The standard deviation of the total amount of time in minutes the operator will spend on the calls each day is of 17.4356 minutes.
0.3085 = 30.85% approximate probability that the total time spent on the calls will be less than 166 minutes.
The value c such that the approximate probability that the total time spent on the calls each day is less than c is 0.95 is [tex]c = 203.4816[/tex]
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
n instances of a normally distributed variable:
For n instances of a normally distributed variable, the mean is:
[tex]M = n\mu[/tex]
The standard deviation is:
[tex]s = \sigma\sqrt{n}[/tex]
Calls to a customer service center last on average 2.3 minutes with a standard deviation of 2 minutes.
This means that [tex]\mu = 2.3, \sigma = 2[/tex]
An operator in the call center is required to answer 76 calls each day.
This means that [tex]n = 76[/tex]
What is the expected total amount of time in minutes the operator will spend on the calls each day?
[tex]M = n\mu = 76*2.3 = 174.8[/tex]
The expected total amount of time in minutes the operator will spend on the calls each day is of 174.8 minutes.
What is the standard deviation of the total amount of time in minutes the operator will spend on the calls each day?
[tex]s = \sigma\sqrt{n} = 2\sqrt{76} = 17.4356[/tex]
The standard deviation of the total amount of time in minutes the operator will spend on the calls each day is of 17.4356 minutes.
What is the approximate probability that the total time spent on the calls will be less than 166 minutes?
This is the p-value of Z when X = 166.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
For this problem:
[tex]Z = \frac{X - M}{s}[/tex]
[tex]Z = \frac{166 - 174.8}{17.4356}[/tex]
[tex]Z = 0.5[/tex]
[tex]Z = 0.5[/tex] has a p-value of 0.6915.
1 - 0.6915 = 0.3085.
0.3085 = 30.85% approximate probability that the total time spent on the calls will be less than 166 minutes.
What is the value c such that the approximate probability that the total time spent on the calls each day is less than c is 0.95?
This is X = c for which Z has a p-value of 0.95, so X = c when Z = 1.645. Then
[tex]Z = \frac{X - M}{s}[/tex]
[tex]1.645 = \frac{c - 174.8}{17.4356}[/tex]
[tex]c - 174.8 = 1.645*17.4356[/tex]
[tex]c = 203.4816[/tex]
The value c such that the approximate probability that the total time spent on the calls each day is less than c is 0.95 is [tex]c = 203.4816[/tex]
The sum of 'n' terms of an arithmetic sequence is 4n^2+3n. What is the first term, the common difference, and the sequence?
Answer:
d=8 and a=7
Step-by-step explanation:
The sum of a arithmetic sequence is given by (n/2)*(2a+(n-1)d). Comparing coefficients with the given Sn, we have; a-d/2=3 and d/2=4, d=8 and a=7. The sequence is 7, 15, 23, 31, 39
The height of the triangle is 4 meters longer than twice its base. find the height if the area of the triangle is 80 square meters. The height must be ___meters
Answer:
The height is 20 meters
Step-by-step explanation:
First set up the equation (Area)80=bh/2 then set up another equation, (height) h=4+2b (base). After this you can substitute h in the equation to end up with 80=(b(2b+4))/2 simplify it to get 80=b^2+2b then solve. The base is 8 meters, plug into the formula that we made before and you find the height is 20 meters.
What is the sum of the infinite geometric series?
Answer:
-6
Step-by-step explanation:
a1= -3
r= -(3/2)/-3 = 0.5
r>-3
s= a1/1-r
= -3/1-0.5
=-6
express 111 as a sum of two primes
Answer:
2 + 109 = 111
Step-by-step explanation:
.............
Convert 2546 in base 10 to base 5
Answer:
40141
Step-by-step explanation:
A particle is moving such that its height h at time t is given by h(t) = 2 + 8t - 3t^2 + 1/5t^3. The average velocity of the particle on the period [0,3] is
[tex]\\ \Large\sf\longmapsto h(t)[/tex]
[tex]\\ \Large\sf\longmapsto 2+8t-3t^2+\dfrac{1}{5}t^3[/tex]
[tex]\\ \Large\sf\longmapsto 2+8(3)-3(3)^2+\dfrac{1}{5}(3)^3[/tex]
[tex]\\ \Large\sf\longmapsto 2+24-3(9)+\dfrac{27}{5}[/tex]
[tex]\\ \Large\sf\longmapsto 26-27+5.4[/tex]
[tex]\\ \Large\sf\longmapsto -2+5.4[/tex]
[tex]\\ \Large\sf\longmapsto h(t)=3.4m[/tex]
please help I have 3 mins left
Answer:
the first one is 3.7 x 10^-4
and the second one is 3.7 x 10^4
explanation:
when we have decimals we are going backward,
therefore "0.00037" would be a negative number
to find the scientific notation form, we have to move the decimal over to the left untill we get 3.7
it took 4 moves to the right to get to 3.7, and since were dealing with decimals it will be negative,
so the first one is 3.7 x 10^4
the second one however is not a decimal so it will be a positive exponent.
now remember that there is always a decimal after a number we might just not see it.
so, going from the very end of the number it takes us 4 moves to the left to get to 3.7
so,
the second one will be 3.7 x 10^4
hope this helped :)
Help on 3,5,7,9,11,13.15,17, please thank you
Answer:
3. 6a+60
5. 25+5w
7. 90-10t
11. 4.5-12c
13. f-2
15. 12z+1.5
Step-by-step explanation:
3.
6(a+10)
Multiply 6 by both factors in the parentheses, in this case, a and 10.
6*a = 6a
6*10 = 60
6(a+10) = 6a + 60
I only put the step- by- step explanation for #3, but you should be able to figure the rest out with that.
Coefficient and degree of the polynomial
Answer:
The leading coefficient is -8 as it is a mix of x and cardinal, if it was x alone then it wouldn't be the coefficient, we would use the next number shown.
If it was just a number and no x then it would still be the coefficient.
The degree is 9 as it is the highest power shown.
Step-by-step explanation:
See attachment for examples
A construction crane lifts a bucket of sand originally weighing 145 lbs at a constant rate. Sand is lost from the bucket at a constant rate of .5lbs/ft. How much work is done in lifting the sand 80ft?
Answer: [tex]10,000\ lb.ft[/tex]
Step-by-step explanation:
Given
Initial weight of the bucket is [tex]145\ lb[/tex]
It is lifted at constant rate and rate of sand escaping is [tex]0.5\ lb/ft[/tex]
At any height weight of the sand is [tex]w(h)=145-0.5h[/tex]
Work done is given by the product of applied force and displacement or the area under weight-displacement graph
from the figure area is given by
[tex]\Rightarrow W=\int_{0}^{80}\left ( 145-0.5h \right )dh\\\\\Rightarrow W=\left | 145h-\dfrac{0.5h^2}{2} \right |_0^{80}\\\\\Rightarrow W=\left [ 145\times 80-\dfrac{0.5(80))^2}{2} \right ]-0\\\\\Rightarrow W=11,600-1600\\\\\Rightarrow W=10,000\ lb.ft[/tex]
HURRY PLEASE!!!!!!
Line AB has a slop of 1/2
What would the slope of line CD have to be if we knew CD was perpendicular to AB?
2
-2
1/2
-1/2
Answer:
-2
Step-by-step explanation:
Perpendicular lines have slopes that are negative reciprocals
Take the slope of AB = 1/2
-1/(1/2)
-1 * 2/1
-2
The slope of a line perpendicular is -2
7. Solve for x: x/6 - y/3 = 1
Please give steps!
2. Solve the following:
a. When six is added to four times a number the result is 50. Find the number.
b. The sum of a number and nine is multiplied by -2 and the answer is -8. Find the
number
10
m in
Step-by-step explanation:
a) let number=x
four times a number=4x
Condition:
4x+6=50
4x=50-6
4x=44
x=44/4
x=11
b) Condition:
x+9×-2=-8
x-18=-8
x=-8+18
x=10
Note:if you need to ask any question please let me know.
What year was it when I was a freshman if I graduated this year(2021)?
Answer:
2019
Step-by-step explanation:
I am assuming you mean graduated from high school. If that's the case it's 2019. Sophmore year means 9th grade. which is 2019!
please help this is due soon
round to the nearest Ten-thousand: 849,708
Answer:
850,000
Step-by-step explanation:
Answer: 850,000
Concept:
Here, we need to know the order and name of each place value.
Please refer to the attachment below for the specified names.
Solve:
8 = Hundred thousands
4 = Ten thousands
9 = One thousands
7 = Hundreds
0 = Tens
9 = Ones
Since the values before the ten thousands place, which would be the one thousands place, is greater than 5, then we should round up.
Therefore, the rounded value would be [tex]\boxed{850,000}[/tex]
Hope this helps!! :)
Please let me know if you have any questions
Need the answer please, soon as possible
9514 1404 393
Answer:
(d) 27.4%
Step-by-step explanation:
The desired percentage is ...
(juniors for Kato)/(total juniors) × 100%
= 129/(129 +194 +147) × 100%
= (129/470) × 100% ≈ 27.4%
About 27.4% of juniors voted for Kato.
Find the area of the shape shown below.
Answer:
28 units²
Step-by-step explanation:
Area of trapezoid =
2(8 + 4)/2 = 12
Area of rectangle =
2 x 8 = 16
16 + 12 = 28
If my answer is incorrect, pls correct me!
If you like my answer and explanation, mark me as brainliest!
-Chetan K
A turboprop plane flying with the wind flew 1,200 mi in 4 h. Flying against the wind, the plane required 5 h to travel the same distance. Find the rate of the wind and the rate of the plane in calm air.
Answer:
30 and 270 respectively
Step-by-step explanation:
Let the speed of plane in still air be x and the speed of wind be y.
ATQ, (x+y)*4=1200 and (x-y)*5=1200. Solving it, we get x=270 and y=30
what is the area of the triangle ://
Answer:
The area of a triangle is:
Area = 1/2(bh)
Area = 1/2(70)
Area = 35 square inches
Let me know if this helps!
Question 17 of 25
Solve the inequality. Enter the answer as an inequality that shows the value of
the variable; for example f>7, or 6 < w. Where necessary, use <= to write s
and use >= to write .
V-(-5) <-9
Answer here
I
SUBMIT
Answer:
v-(-5)<-9
v- remove brackets -5
v- -5= -4 +5 ( opposite operation)
v- = -4
v< -4
What is the dimension of the null space Null (A) of A =
Answer:
the nullity of a matrix A is the demision of its null space:nullity A = dim (n(A).
The figure shows an equilateral triangle with its sides as indicated. find the length of each side of the triangle .
I Will Mark Brainliest
Answer:
21
Step-by-step explanation:
All three sides are equal
2x-7 = x+y-9 = y+5
Using the last two
x+y-9 = y+5
Subtract y from each side
x+y-9-y = y+5-y
x-9 = 5
Add 9 to each side
x -9+9 = 5+9
x=14
We know the side length is
2x-7
2(14) -7
28-7
21
The side length is 21
What is the volume of the cylinder below?
Height 4
Radius 7
Answer:
V ≈ 615.75
r Radius 7
h Height 4
Can someone please help me with this math problem
We have [tex]f\left(f^{-1}(x)\right) = x[/tex] for inverse functions [tex]f(x)[/tex] and [tex]f^{-1}(x)[/tex]. Then if [tex]f(x) = 2x+5[/tex], we have
[tex]f\left(f^{-1}(x)\right) = 2f^{-1}(x) + 5 = x \implies f^{-1}(x) = \dfrac{x-5}2[/tex]
Then
[tex]f^{-1}(8) = \dfrac{8-5}2 = \boxed{\dfrac32}[/tex]