Answer:
Maximum speed of bird A is [tex]192\,\,\frac{mi}{h}[/tex]
Maximum speed of bird B is [tex]32\,\,\frac{mi}{h}[/tex]
Step-by-step explanation:
This is a problem with two unknowns: Max speed of bird A (we name that "A"), and max speed of bird B (we call that "B"). Now we can create two equations with these two unknowns, based on the info provided:
Equation 1): based on the phrase "bird A can travel six times as fast as bird B" we write:
[tex]A=6\,*\, B\\A=6B[/tex]
Equation 2): based on the phrase; "the total speeds for these two birds is 224 miles per hour", we write:
[tex]A+B=224\,\,\frac{mi}{h}[/tex]
Now, we use the first equation to substitute A in the second equation, ad then solve for the unknown B:
[tex]A+B=224\,\,\frac{mi}{h}\\(6B)+B=224\,\,\frac{mi}{h}\\7B=224\,\,\frac{mi}{h}\\B=\frac{224}{7} \,\,\frac{mi}{h}\\B=32\,\,\frac{mi}{h}[/tex]
Now we can solve for the other unknown "A" using the substitution equation and the value of B we just found:
[tex]A=6B\\A=6\,(32\,\,\frac{mi}{h})\\A=192\,\,\frac{mi}{h}[/tex]
2.In a large university 13.5% of the students take economics, 24.7% of the students take statistics and 11.7% take economics and statistics. The probability that a randomly selected student didn’t take economics but did take statistics is close toالقارئ الشامل (2/2 نقط
Answer:
The probability that a randomly selected student didn’t take economics but did take statistics is 13%.
Step-by-step explanation:
Let the event that a student offers Economics be E.
The event that a student does NOT offer Economics is E'.
Let the event that a student offers Statistics be S.
The event that a student does NOT offer Statistics be S'.
P(E) = 13.5% = 0.135
P(S) = 24.7% = 0.247
P(E n S) = 11.7% = 0.117
Find the probability that a randomly selected student didn’t take economics but did take statistics
This probability = P(E' n S)
Since E and E' are mutually exclusive events,
P(S) = P(E' n S) + P(E n S)
P(E' n S) = P(S) - P(E n S)
P(E' n S) = 0.247 - 0.117 = 0.13 = 13%
Hope this Helps!!!
In general, the probability that a blood donor has Type A blood is 0.40.Consider 8 randomly chosen blood donors, what is the probability that more than half of them have Type A blood?
The probability that more than half of the 8 randomly chosen blood donors have Type A blood is approximately 0.2533 or 25.33%.
To calculate the probability that more than half of the 8 randomly chosen blood donors have Type A blood, we can use the binomial probability formula:
[tex]\mathrm{P(X > n/2) = \sum [ P(X = k) ]}[/tex]
where the sum is taken from k = (n/2 + 1) to k = n
In this case, n represents the number of trials (8 blood donors) and p is the probability that a single blood donor has Type A blood (0.40).
P(X = k) is the probability of getting exactly k donors with Type A blood, and it is given by the binomial probability formula:
[tex]\mathrm {P(X = k) = (n, k) \times p^k \times (1 - p)^{(n - k)}}[/tex]
where (n choose k) represents the number of combinations of n items taken k at a time, and it is given by:
[tex]\mathrm {(n, k) = \frac{n!}{(k! \times (n - k)!)}}[/tex]
Now, let's calculate the probability that more than half (i.e., 5 or more) of the donors have Type A blood:
[tex]\mathrm{P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)}[/tex]
[tex]\mathrm {P(X = k) = (8, k) \times 0.40^k \times (1 - 0.40)^{(8 - k)}}[/tex]
[tex]\mathrm{P(X = 5)} = (8, 5) \times 0.40^5 \times (1 - 0.40)^{(8 - 5)}\\\\= 56 \times 0.01024 \times 0.343\\\\= 0.1961984[/tex]
[tex]\mathrm{P(X = 6)} = (8, 6) \times 0.40^6 \times (1 - 0.40)^{(8 - 6)}\\\\= 28 \times 0.004096 \times 0.36\\\\= 0.0516608[/tex]
[tex]\mathrm {P(X = 7)} = (8, 7) \times 0.40^7 \times (1 - 0.40)^{(8 - 7)}\\\\= 8 \times 0.0016384 \times 0.4\\\\= 0.0052224[/tex]
[tex]\mathrm {P(X = 8)} = (8, 8) \times 0.40^8 \times (1 - 0.40)^{(8 - 8)}\\\\= 1 \times 0.00065536 \times 0.4\\\\= 0.000262144[/tex]
Now, add all these probabilities together to get the final result:
[tex]\mathrm {P(X > 4)} = 0.1961984 + 0.0516608 + 0.0052224 + 0.000262144\\\\= 0.253343344[/tex]
Therefore, the probability that more than half of the 8 randomly chosen blood donors have Type A blood is approximately 0.2533 or 25.33%.
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In a grinding operation, there is an upper specification of 3.150 in. on a dimension of a certain part after grinding. Suppose that the standard deviation of this normally distributed dimension for parts of this type ground to any particular mean dimension LaTeX: \mu\:is\:\sigma=.002 μ i s σ = .002 in. Suppose further that you desire to have no more than 3% of the parts fail to meet specifications. What is the maximum (minimum machining cost) LaTeX: \mu μ that can be used if this 3% requirement is to be met?
Answer:
Step-by-step explanation:
Let X denote the dimension of the part after grinding
X has normal distribution with standard deviation [tex]\sigma=0.002 in[/tex]
Let the mean of X be denoted by [tex]\mu[/tex]
there is an upper specification of 3.150 in. on a dimension of a certain part after grinding.
We desire to have no more than 3% of the parts fail to meet specifications.
We have to find the maximum [tex]\mu[/tex] such that can be used if this 3% requirement is to be meet
[tex]\Rightarrow P(\frac{X- \mu}{\sigma} <\frac{3.15- \mu}{\sigma} )\leq 0.03\\\\ \Rightarrow P(Z <\frac{3.15- \mu}{\sigma} )\leq 0.03\\\\ \Rightarrow P(Z <\frac{3.15- \mu}{0.002} )\leq 0.03[/tex]
We know from the Standard normal tables that
[tex]P(Z\leq -1.87)=0.0307\\\\P(Z\leq -1.88)=0.0300\\\\P(Z\leq -1.89)=0.0293[/tex]
So, the value of Z consistent with the required condition is approximately -1.88
Thus we have
[tex]\frac{3.15- \mu}{0.002} =-1.88\\\\\Rrightarrow \mu =1.88\times0.002+3.15\\\\=3.15[/tex]
SELECT THE EQUIVALENT EXPRESSION
(6^-4 x 8^-7)^-9
A. 6^36•8^63
B. 1/6^13•8^16
Answer:
A
Step-by-step explanation:
Calculate the products in the multiple choice and see if any equal the product in the problem.
Hence as the products calculated in choice A equal that in the problem;the answer is A
Each limit represents the derivative of some function f at some number a. State such an f and a in each case.
lim √9 + h - 3 / h
h-->0
Answer:
a = 0f(h) = [tex]\frac{\sqrt{9+h} - 3}{h}[/tex]limit of the function is 1/6Step-by-step explanation:
The general form representing limit of a function is expressed as shown below;
[tex]\lim_{h \to a} f(h)[/tex] where a is the value that h will take and use in the function f(h). It can be expressed in words as limit of function f as h tends to a. Comparing the genaral form of the limit to the limit given in question [tex]\lim_{h \to 0} \frac{\sqrt{9+h} - 3}{h}[/tex], it can be seen that a = 0 and f(h) = [tex]\frac{\sqrt{9+h} - 3}{h}[/tex]
Taking the limit of the function
[tex]\lim_{h \to 0} \frac{\sqrt{9+h} -3}{h}\\= \frac{\sqrt{9+0}-3 }{0}\\= \frac{0}{0}(indeterminate)[/tex]
Applying l'hopital rule
[tex]\lim_{h \to 0} \frac{\frac{d}{dh} (\sqrt{9+h} - 3)} {\frac{d}{dh} (h)}\\= \lim_{h \to 0} \frac{1}{2} (9+h)^{-1/2} /1\\=\frac{1}{2} (9+0)^{-1/2}\\= \frac{1}{2} * \frac{1}{\sqrt{9} } \\= 1/2 * 1/3\\= 1/6[/tex]
The mean family income for a random sample of 600 suburban households in Loganville shows that a 95 percent confidence interval is ($43,100, $59,710). Alma is conducting a test of the null hypothesis H0: µ = 42,000 against the alternative hypothesis Ha: µ ≠ 42,000 at the α = 0.05 level of significance. Does Alma have enough information to conduct a test of the null hypothesis against the alternative?
Answer:
[tex] 43100 \leq \mu \leq 59710[/tex]
And for this case we want to test the following hypothesis:
Null hypothesis: [tex] \mu =42000[/tex]
Alternative hypothesis: [tex] \mu \neq 42000[/tex]
For this case since the lower value of the confidence interval is higher than 42000 we have enough evidence to reject the null hypothesis at the 55 of significance and we can conclude that the true mean is significantly different from 42000
Step-by-step explanation:
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
And for this case the 95% confidence interval is already calculated as:
[tex] 43100 \leq \mu \leq 59710[/tex]
And for this case we want to test the following hypothesis:
Null hypothesis: [tex] \mu =42000[/tex]
Alternative hypothesis: [tex] \mu \neq 42000[/tex]
For this case since the lower value of the confidence interval is higher than 42000 we have enough evidence to reject the null hypothesis at the 55 of significance and we can conclude that the true mean is significantly different from 42000
Answer: Yes, because $42,000 is not contained in the 95% confidence interval, the null hypothesis would be rejected in favor of the alternative, and it could be concluded that the mean family income is significantly different from $42,000 at the α = 0.05 level
Step-by-step explanation:
took the test
I NEED HELP WITH THIS PLEASE HELP ME
Answer:
156 minutes
Step-by-step explanation:
So we need to create an equation to represent how Frank's phone company bills him
I will denote "y" as the total for his billI will denote "x" as the number of minutes Frank usesSo the phone company charges an $8 monthly fee, so this value remains constant and will be our "y-intercept"
They then charge $0.06 for every minute he talks, this will be our "slope"
Combining everything into an equation, we have: y = 0.06x + 8
Now since we were given Franks phone bill total and want to figure out how many minutes he used, we just need to solve the equation for x and plug in our known y value
y = 0.06x + 8 → y - 8 = 0.06x → [tex]x=\frac{y-8}{0.06}[/tex] Then plugging in our y value we get [tex]x=\frac{17.36-8}{0.06}=\frac{9.36}{0.06}= 156[/tex]Frank used up a total of 156 minutes
For a super soaker water gun, a pump handle is moved back and forth to build up pressure in the water reservoir. The water is released by pulling a trigger and shooting the water a significant distance. Assuming that you can create an absolute pressure of 8 atm in the reservoir:
a) What is the velocity at which the water leaves the gun?
b) If the water exits the gun through a hole with a radius of 1-mm, what is the volume rate of flow in m3/s?
c) If the water gun is fired horizontally and held 1.2 meters above the ground, where does the water hit the ground? Pressure 8 cm water
Answer:
a) The velocity at which the water leaves the gun = 37.66 m/s
b) The volume rate of flow = (1.183 × 10⁻⁴) m³/s
c) The water hits the ground 18.64 m from the point where the water gun was shot.
Step-by-step explanation:
a) Using Bernoulli's equation, an equation that is based on the conservation of energy.
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
The two levels we are considering is just inside the water reservoir and just outside it.
ρgh is an extension of potential energy and since the two levels are at the same height,
ρgh₁ = ρgh₂
Bernoulli's equation becomes
P₁ + ½ρv₁² = P₂ + ½ρv₂²
P₁ = Pressure inside the water reservoir = 8 atm = 8 × 101325 = 810,600 Pa
ρ = density of water = 1000 kg/m³
v₁ = velocity iof f water in the reservoir = 0 m/s
P₂ = Pressure outside the water reservoir = atmospheric pressure = 1 atm = 1 × 101325 = 101,325 Pa
v₂ = velocity outside the reservoir = ?
810,600 + 0 = 101,325 + 0.5×1000×v₂²
500v₂² = 810,600 - 101,325 = 709,275
v₂² = (709,275/500) = 1,418.55
v₂ = √(1418.55) = 37.66 m/s
b) Volumetric flowrate is given as
Q = Av
A = Cross sectional Area of the channel of flow = πr² = π×(0.001)² = 0.0000031416 m²
v = velocity = 37.66 m/s
Q = 0.0000031416 × 37.66 = 0.0001183123 m³/s = (1.183 × 10⁻⁴) m³/s
c) If the height of gun above the ground is 1.2 m. Where does the water hit the ground?
The range of trajectory motion is given as
R = vT
v = horizontal component of the velocity = 37.66 m/s
T = time of flight = ?
But time of flight is given as
T = √(2H/g) (Since the initial vertical component of the velocity = 0 m/s
H = 1.2 m
g = acceleration due to gravity = 9.8 m/s²
T = √(2×1.2/9.8) = 0.495 s
Range = vT = 37.66 × 0.495 = 18.64 m
Hope this Helps!!!
Assume that in a statistics class the probability of receiving a grade of A equals .30 and the probability of receiving a grade of B equals .30. The probability that a randomly selected student from this class will receive either an A or a B equals.
a. .09
b. .6
c. .9
d. .3
Answer:
Answer D is correct
Please answer this question for me thank you !! 20 Points !! Will give brainliest !!
Answer:
b
Step-by-step explanation:
In a parralel graph, the slopes would always be the same. The intercept in the answer is 2, showing that the coordinate points are (0,2)
Hope this helps!:)
Answer:
B) y = 2x + 2
Step-by-step explanation:
Firstly, you have to know that parallel lines have congruent slopes. That means that the slope of this line will be 2.
Next, make a point slope form of the equation:
y - y1 = m(x - x1)
y - 2 = 2(x - 0)
y - 2 = 2x - 0
Now, we can make it into slope intercept form.
y - 2 = 2x
y = 2x + 2
Hope this helps :)
(Please hurry)
Explain how to find the value of x
Answer:
96
Step-by-step explanation:
Exterior angles add up to 360
360 - 134-130 = 96
x = 96
Nam owns a used car lot. He checked the odometers of the cars and recorded how far they had driven. He
then created both a histogram and a box plot to display this same data (both diagrams are shown below).
Which display can be used to find how many vehicles had driven more than 200,000 km (kilometers)?
Choose 1 answer:
Answer:
a histogram
Step-by-step explanation:
You can count easily from hiistogram how many vehicles had driven more than 200,000 km (kilometers) and that's not the case with the box plot
What’s the correct answer for this question?
Answer: choice D 1/2
Step-by-step explanation:
Events A and B are independent if the equation P(A∩B) = P(A) · P(B) holds true.
so
1/6=1/3*p(A)
p(A)=1/2
How do I set up this problem. I'm lost
Answer:
the answer is 64 .
Step-by-step explanation:
basically i just divided 48 by 2.4 and got 20 .. so that means that 20 has to be the multiplied factor so i just multiplied 3.2 by 20 and got 64.
if y=5x what happens to the value of y if the value of x doubles
Answer:
[tex] y = 5x[/tex]
And we need to ee what happen if we increase the value of x by a factor of 2. So then for this case we can set up the equation like this:
[tex] y_f = 5(2x) = 10x[/tex]
And if we find the ratio between the two equations we got:
[tex] \frac{y_f}{y} =\frac{10x}{5x} =2[/tex]
So then if we increase the value of x by a factor of 2 then the value of y increase also by a factor of 2
Step-by-step explanation:
For this case we have this equation given:
[tex] y = 5x[/tex]
And we need to ee what happen if we increase the value of x by a factor of 2. So then for this case we can set up the equation like this:
[tex] y_f = 5(2x) = 10x[/tex]
And if we find the ratio between the two equations we got:
[tex] \frac{y_f}{y} =\frac{10x}{5x} =2[/tex]
So then if we increase the value of x by a factor of 2 then the value of y increase also by a factor of 2
An object travels along a horizontal path at a constant rate.the object travels 1/20 of the length of the path in 3/4 second.at that rate,how many seconds does it take the object to travel the entire length of the path?
Answer:
The onject 1/8 of the length of the path 3/4 in second.
Using the ratio and proportion to find the total time does it take the object to travel the entire length of the path as following
Length:time
X:(total time )
Total time x.(3/4)/(1/8x)=(3/4)/(1/8) = 6 seconds
Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications, which if a multiplication takes sec would take these times:
n 10 15 20 25
Time 0.004 sec 22 min 77 years 0.5.109years
Answer:
number of multiplies is n!n=10, 3.6 msn=15, 21.8 minn=20, 77.09 yrn=25, 4.9×10^8 yrStep-by-step explanation:
Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...
mpy[n] = n·mp[n-1]
mpy[2] = 2
So, ...
mpy[n] = n! . . . n ≥ 2
__
If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...
10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10
Then the larger matrices take ...
n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min
n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years
n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years
_____
For the shorter time periods (less than 100 years), we use 365.25 days per year.
For the longer time periods (more than 400 years), we use 365.2425 days per year.
In a random sample of high school seniors, the proportion who use text messaging was 0.88. In a random sample of high school freshmen, this proportion was 0.68. Researchers found the difference in proportions to be statistically significant and obtained one of the following numbers for the p-value. Which is it?
a. 1.5
b. 0.02
c. 0.78
d. 0.30
Answer:
b. 0.02
Step-by-step explanation:
The smaller the p-value, the stronger the evidence that you should reject the null hypothesis. In this case, this will mean rejecting that the proportions are not significantly different.
Usually, a p-value is considered to be statistically significant when p ≤ 0.05.
From the answer options provided, alternative b. 0.02 is the only one that represents the difference in proportions to be statistically significant (there is only a 2% chance that the proportions are not significantly different).
Therefore, the answer is b. 0.02
please hurry I’ll make brainiest
A marble is thrown off of a balcony, towards the ground, from a height
of 18 feet above ground level, with a velocity of 4.5 feet per second.
Which function could be used to model the height of the marble, after
t seconds?
Answer:
Option (3)
Step-by-step explanation:
A stone has been thrown off towards the ground from a height [tex]h_{0}[/tex] = 18 feet
Initial speed of the stone 'u' = 4.5 feet per second
Since height 'h' of a projectile at any moment 't' will be represented by the function,
h(t) = ut - [tex]\frac{1}{2}(g)(t)^2[/tex] + [tex]h_{0}[/tex]
h(t) = 4.5t - [tex]\frac{1}{2}(32)t^2[/tex]+ 18 [ g = 32 feet per second square]
h(t) = 4.5t - 16t² + 18
h(t) =-16t² + 4.5t + 18
Therefore, Option (3) will be the answer.
Suppose ARB Bank is reviewing its service charges and interest payment policies on current accounts. Suppose further that ARB has found that the average daily balance on personal current accounts is GH¢350.00, with a standard deviation of GH¢160.00. In addition, the average daily balances have been found to follow a normal distribution;
What percentage of customers carries a balance of GH¢100 or lower?
What percentage of customers carries a balance of GH¢500 or lower?
What percentage of current account customers carries average daily balances exactly equal to GH¢500?
What percentage of customers maintains account balance between GH¢100 and GH¢500?
Answer:
5.94% of customers carries a balance of GH¢100 or lower.
82.64% of customers carries a balance of GH¢500 or lower.
0% of current account customers carries average daily balances exactly equal to GH¢500.
76.7% of customers maintains account balance between GH¢100 and GH¢500
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 350, \sigma = 160[/tex]
What percentage of customers carries a balance of GH¢100 or lower?
This is the pvalue of Z when X = 100. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{100 - 350}{160}[/tex]
[tex]Z = -1.56[/tex]
[tex]Z = -1.56[/tex] has a pvalue of 0.0594
5.94% of customers carries a balance of GH¢100 or lower.
What percentage of customers carries a balance of GH¢500 or lower?
This is the pvalue of Z when X = 500.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{500 - 350}{160}[/tex]
[tex]Z = 0.94[/tex]
[tex]Z = 0.94[/tex] has a pvalue of 0.8264
82.64% of customers carries a balance of GH¢500 or lower.
What percentage of current account customers carries average daily balances exactly equal to GH¢500?
In the normal distribution, the probability of finding a value exactly equal to X is 0. So
0% of current account customers carries average daily balances exactly equal to GH¢500.
What percentage of customers maintains account balance between GH¢100 and GH¢500?
This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 100.
From b), when X = 500, Z = 0.94 has a pvalue of 0.8264
From a), when X = 100, Z = -1.56 has a pvalue of 0.0594
0.8264 - 0.0594 = 0.767
76.7% of customers maintains account balance between GH¢100 and GH¢500
You have $150 to spend at a store. If you shoes cost $30 and belts cost $25, write an equation that represents the different ways that you could spend a total of $150
Answer:
you could buy a pair of shoes and a belt still have 95 dollars to spend
Simplify this equation x2-5x-36
Answer:
[tex]=\left(x+4\right)\left(x-9\right)[/tex]
Step-by-step explanation:
[tex]x^2-5x-36\\\mathrm{Break\:the\:expression\:into\:groups}\\=\left(x^2+4x\right)+\left(-9x-36\right)\\\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2+4x\mathrm{:\quad }x\left(x+4\right)\\\mathrm{Factor\:out\:}-9\mathrm{\:from\:}-9x-36\mathrm{:\quad }-9\left(x+4\right)\\=x\left(x+4\right)-9\left(x+4\right)\\\mathrm{Factor\:out\:common\:term\:}x+4\\=\left(x+4\right)\left(x-9\right)[/tex]
If the discriminant of a quadratic equation is equal to -8, which statement describes the roots?
There are two complex roots.
There are two real roots.
There is one real root.
There is one complex root.
Answer:
There are two complex roots.
Step-by-step explanation:
When the discriminant is a negative number, the parabola will not intersect the x-axis. This means that there are no solutions/two complex solutions.
Isaac is organizing a 5-kilometer road race. The safety committee
recommends having a volunteer every 1 of a kilometer and at
the finish.
| Are 10 volunteers enough?
Answer:
10 volunteers are more than recommendedStep-by-step explanation:
The recommended number of volunteers is five (5)
Since the the distance of the race is 5km,
and the safety committees recommends 1 volunteer per kilometre.
Hence ten (10) volunteers is more than enough
PLS HELP ME 10PTS
An artist creates a cone-shaped sculpture for an art exhibit. If the sculpture is 7 feet tall and has a base with a circumference of 27.632 feet, what is the volume of the sculpture?
Answer: The volume of the sculpture is 141.84 cubic-feet
Step-by-step explanation: Please see the attachments below
Which triangle’s area would be calculated using the trigonometric area formula?
Triangle E F D is shown. The length of E F is 10, the length of D F is 7, and the length of D E is 12.
Triangle Q R P is shown. The length of Q R is 5 and the length of R P is 6. Angle Q R P is 40 degrees.
Triangle A B C is shown. The length of A B is 4 and the length of B C is 5. Angle B C A is 25 degrees.
Triangle X Y Z is shown. The length of Y Z is 4. Angle Z X Y is 29 degrees and angle X Y Z is 110 degrees.
Answer:
Triangle Q R P is shown. The length of Q R is 5 and the length of R P is 6. Angle Q R P is 40 degrees.
Step-by-step explanation:
The trigonometric formula refers the two sides length of the triangle and it also consists of included angle to find out the area
A = [tex]\frac{1}{2}[/tex] ab sin C
QPR contains two sides and the included angle
XYZ has one side and the two angles
DEF has only three sides
And, the ABC contains two sides but does not have the included angle
Based on the explanation above, the correct option is B
Answer: the second option aka B
Step-by-step explanation: The other person explained it and I'm just here to tell you they gave the correct and answer for edge 2020.
17)Let f(x) = -2x + 5 and g(x) = 9x2 + 4. Find f(8) + g(8) . A)565 B)569 C)564 D)560
Answer:
answer B [tex]\boxed{ \ 569 \ }\\[/tex]
Step-by-step explanation:
f(8)=-2*8+5=-11
g(8)=9*8*8+4=580
f(8)+g(8)= -11+580=569
if you’re good with permutations in math 30 help out with this easy question
In how many ways can five boys and three girls sit in a row such that all boys sit together?
a) 4800
b) 5760
c) 2880
d) 1440
Answer:
2880
Step-by-step explanation:
Consider the 5 boys to be 1 group. The boys and 3 girls can be arranged in 4! ways.
Within the group, the boys can be arranged 5! ways.
The total number of permutations is therefore:
4! × 5! = 2880
If a triangle has sides that are 21 and 6 what is the range for third side x?
Enter your answer without spaces in range format.
Example: 1<x<3
Answer:
15<x<27
Step-by-step explanation:
Rule for the sides of a triangle:
The sum of the two smallest sides of a triangle must be greater than the biggest side.
In this question:
Sides of 6, 21 and x. We have to find the range for x.
If 21 is the largest side:
Two smallest are 6 and x.
x + 6 > 21
x > 21 - 6
x > 15
If x is the largest side:
Two smallest and 6 and 21. So
21 + 6 > x
27 > x
x < 27
Then
x has to be greater than 15 and smaller than 27. So the answer is:
15<x<27
a kangaroo and a wallaby are in a race. They have to get to a flagbole that is 100 meters away and back. For every 2 hops the kangaroo does, the wallaby does three but the kangaroo's jumps are 3 meters while the wallaby's are 2. Who gets there and back first (hint: it isnt a draw)
Answer:
im going to say a wallaby because they are smaller and lighter and if you think of the weight then less power is needed for a wallaby
idk lol XD
Step-by-step explanation: