The heights of fully grown English oak (also known as Brown oak) trees are normally distributed with a mean of 95 feet and a standard deviation of 10 feet. What proportion of fully grown English oak trees are taller than 110 feet?
Final answer:
Using the z-score calculation and the standard normal distribution, approximately 6.68% of fully grown English oak trees are taller than 110 feet.
Explanation:
To find the proportion of fully grown English oak trees taller than 110 feet, given that the heights are normally distributed with a mean of 95 feet and a standard deviation of 10 feet, we first calculate the z-score for a height of 110 feet. The z-score is calculated using the formula: z = (X - μ) / σ, where X is the value of interest (110 feet), μ is the mean (95 feet), and σ is the standard deviation (10 feet).
By substituting the given values: z = (110 - 95) / 10 = 1.5. This z-score represents the number of standard deviations 110 feet is above the mean. To find the proportion of trees taller than 110 feet, we consult the standard normal distribution table or use a calculator that provides the area to the right of the z-score, which corresponds to the proportion we seek.
The standard normal distribution table or a calculator shows that the proportion of the area under the curve to the right of a z-score of 1.5 is approximately 0.0668.
Thus, about 6.68% of fully grown English oak trees are taller than 110 feet.
Steven has 14 steel balls of equal weight. If he puts 9 of them in one pan of a pan balance and the rest along with a weight of 20 grams in the other pan, the pans balance each other. What is the weight of one steel ball?
Answer:
Step-by-step explanation:
How many solutions are there to the system of equations graphed below if the lines are parallel
Answer:
Zero
C is the correct option.
Step-by-step explanation:
In the given graph, the given lines are parallel to each other and hence never intersect.
Now, we know that the intersection point (s) of two curves gives us the solution.
Here the lines are parallel to each other and hence do not intersect each other.
Thus, there would not be any intersection point.
Hence, the system of equations has no solution.
C is the correct option.
14m+m+17m−9m if m=30
Answer:
690
Step-by-step explanation:
Put 30 where m is in the expression and do the arithmetic. Or, simpify the expression first, then do the same.
14·30 +30 +17·30 -9·30 = 420 +30 +510 -270 = 690
Simplifying first, you have ...
14m +m +17m -9m = m(14 +1 +17 -9) = 23m
23·30 = 690
To solve the equation 14m + m + 17m - 9m when m=30, substitute the value of m in the equation and follow order or operations. The result is 960.
Explanation:To solve for the given problem involving the variable m, simply substitute the value of m into the equation. The equation given is 14m + m + 17m - 9m. Substituting 30 for m in the equation gives us: 14(30) + 30 + 17(30) - 9(30). To solve this, use the order of operations principle, which states that multiplication and division should be performed before addition and subtraction.
First, perform the multiplication: 420 + 30 + 510 - 270. Next, add and subtract from left to right: 960. Thus, if m = 30, then 14m + m + 17m - 9m = 960.
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Andrew is given a weekly allowance of $17.50 he uses $4.50 for lunch on Monday what amount at the most can he spend each day if he wants to spend the same amount each of the remaining 4 days write an inequality and solve
The inequality representing Andrew's spending limit is d <= $3.25, where d is the daily spending amount for the remainder of the week. Solving this, Andrew can spend at most $3.25 per day for the next 4 days.
Andrew received a weekly allowance of $17.50. He spent $4.50 on lunch on Monday. The question is how much at most can he spend each day for the remaining 4 days if he spends the same amount each day. To answer this, we subtract the amount spent on Monday from the total weekly allowance to find the remaining balance. Then, we divide this remaining balance by the number of days left in the week to find the daily spending limit:
Total allowance = $17.50
Spent on Monday = $4.50
Remaining balance = Total allowance - Spent on Monday
= $17.50 - $4.50
= $13.00
Days left in the week = 4
Daily spending limit = Remaining balance \/ Days left
= $13.00 / 4
= $3.25
An American car company has designed a new high fuel efficiency vehicle that is rated at 55 miles per gallon. The company plans to export the car to Europe and must advertise the fuel efficiency in SI units. What is the fuel usage rate in kilometers per liter?
8x-2=46 what is this
based only on the information given in the diagram, which congruence theorems or postulates could be given as reasons why ΔLMN ≈ΔOPQ?
Check all that apply.
A. LL
B. LA
C. SAS
D. HL
E. ASA
F. AAS
HL congruence theorems or postulates used to prove ΔLMN ≈ΔOPQ.
What is Congruency?Congruent figures have sides that are the same length and angles that are the same measurement. In other words, congruent figures are those in which one figure superimposes another.
If two line segments have the same length, they are said to be congruent. If two angles have the same measure, they are said to be congruent. If the matching sides and angles of two triangles are equal, they are said to be congruent.
We have two Triangles as ΔLMN and ΔOPQ
As, per the figure both triangles are Right angles Triangle.
Also, LM = OP
MN = PQ
LN = OQ
and, the hypotenuse of both triangles also Equal.
Thus, both triangle are Congruent using HL theorem.
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How can I solve this please:
Solve for q: (4q - 6n)/(3m) = f
General Idea:
Solving for a variable means getting that variable by itself by UNDOING whatever is done to it. We need to perform reverse operation to UNDO.
Applying the concept:
[tex] \frac{4q-6n}{3m}=f\\ \\ To \; undo \; 3m\; in\; the \; denominator \; 3m, \; MULTIPLY \; 3m\; on\; both\; sides [/tex]
[tex] 3m \times \frac{4q-6n}{3m}=f \times 3m\\ \\ 4q-6n=3mf\\Add \; 6n \; both\; sides\\ \\ 4q-6n+6n=3mf+6n\\ Combine \; like\; terms\\ \\ 4q=3mf+6n\\Divide \; 4\; on\; both\; sides\\ \\ \frac{4q}{4} =\frac{3mf+6n}{4} \\ Simplify\; fraction\; in\; left\; sides\\ \\ q=\frac{3mf+6n}{4} [/tex]
Conclusion:
[tex] Solving \; the \; equation \; \frac{4q-6n}{3m}=f \; for q, \; we get...\\ \\q=\frac{3mf+6n}{4} [/tex]
Find the sum of the first 30 terms of the sequence. a_(n)=4n+1
A certain car costs $11,595 before taxes are added. Taxes are $860 and license tags cost $95. What is the overall tax rate (to the nearest tenth)?
0.8%
7.4%
8.2%
12.1%
Answer:
8.2%
Step-by-step explanation:
o-ware
NEED HELP ASAP:
Suppose you pay $24 for a pair of shoes that has been discounted 20%. What is the original price of the shoes? Show how you identify what you are looking for, set up an equation, arrive at your answer, and check your work. Then clearly state your answer.
On average how many words can 400 turtles type in 400 minutes?
Can you guys help me figure this out ?
Is the figure on the right congruent to the sample figure? Explain.
write the equation of the line with the following chracteristics
perpendicular to y=1/2x+3 through (9,1)
A basketball player makes 40% of his shots from the free throw line. suppose that each of his shots can be considered independent and that he throws 3 shots. let x = the number of shots that he makes. what is the probability that he makes 2 shots or less?
Using the binomial distribution, it is found that there is a 0.936 = 93.6% probability that he makes 2 shots or less.
-------------------
For each throw, there are only two possible outcomes. Either the player makes it, or he misses it. The probability of making a throw is independent of any other throw, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
x is the number of successes.n is the number of trials.p is the probability of a success on a single trial.In this question:
The player makes 40% of the shots, thus [tex]p = 0.4[/tex]3 shots, thus [tex]n = 3[/tex]The probability of making 2 or less is the probability that he does not make all of them, that is:
[tex]P(X < 3) = 1 - P(X = 3)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{3,3}.(0.4)^{3}.(0.6)^{0} = 0.064[/tex]
[tex]P(X < 3) = 1 - P(X = 3) = 1 - 0.064 = 0.936[/tex]
0.936 = 93.6% probability that he makes 2 shots or less.
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bonnie had 6 bags of 6 gel pens. she wanted to give the same number of gel pens to 9 friends. how many gel pens did each friend get?
Cheryl falkowski has a collection of silver spoons from all over the world. she finds that she can arrange her spoons in sets of 77 with 44 left over, sets of 88 with 33 left over, or sets of 1515 with 1414 left over. if cheryl has fewer than 200 spoons, how many are there?
Brett is making a fruit salad. The recipe calls for 1
1
2
cups of apple,
3
4
cup of oranges, and
2
3
cup of grapes. How many cups of fruit salad will Brett’s recipe make
A family has two cars. The first car has a fuel efficiency of 35 miles per gallon of gas and the second has a fuel efficiency of 20 miles per gallon of gas. During one particular week, the two cars went a combined total of 2025 miles, for a total gas consumption of 75 gallons. How many gallons were consumed by each of the two cars that week?
Find the orbital period (in years) of an asteroid whose average distance from the sun is 10 au.
The orbital period (in years) of an asteroid whose average distance from the sun is 10 AU is 212314723 years.
What is Orbital period ?
Orbital period of a body is the the time taken by a body to cover one circular path around the central object under whose influence it is following circular path.
Given is a asteroid moving around the sun.
Mean distance from the Sun [x] = 10 AU = 1.496 x 10¹² m.
Assume that the time period of the asteroid is T.
The formula to calculate the orbital period around the sun is -
T²/ R³ = 4π²/GM[s]
Where -
R is the mean distance from sun.
G is the universal Gravitation constant.
M[s] is the mass of Sun.
Substituting the values, we get -
T² = (1.496 x 10¹² x 4 x 3.14 x 3.14)/(6.673 x 10⁻¹¹ x 1.98 x 10³⁰)
T² = (59 x 10¹²)/(13.2 x 10⁻¹⁹)
T² = 4.45 x 10 x 10³⁰
T² = 44.5 x 10³⁰
T = √44.5 x √10³⁰
T = 6.7 x 10¹⁵ seconds
In years, the time period will be -
T = 212314723 years (approx.)
Therefore, the orbital period (in years) of an asteroid whose average distance from the sun is 10 AU is 212314723 years.
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Officer Brimberry wrote 8 tickets for traffic violations last week, but only 7 tickets this week. What is the percent decrease? Give your answer to the nearest tenth of a percent.
The value of √222 is between which two integers?
a) between 13 and 14
b) between 14 and 15
c) between 15 and 16
d) between 12 and 13
what is a solution to the equation 6t=114
Evaluate these quantities.
a.−17 mod 2
b.144 mod 7
c.−101 mod 13
d.199 mod 19
The modulo operator determines the remainder of a division operation. For the given expressions, the results are a. 1, b. 4, c. 2, d. 6
Explanation:The question is asking to evaluate several expressions using the modulo operation. Modulo is a mathematical operation that finds the remainder or signed remainder of a division, after one number is divided by another (called the modulus). Here's how we can solve each:
a. −17 mod 2 = 1 (Because when -17 is divided by 2, the remainder is 1) b. 144 mod 7 = 4 (Because when 144 is divided by 7, the remainder is 4) c. −101 mod 13 = 2 (Because when -101 is divided by 13, the remainder is 2) d. 199 mod 19 = 6 (Because when 199 is divided by 19, remainder is 6)Learn more about Modulo Operation here:https://brainly.com/question/36617304
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Determine (without solving the problem) the maximal interval in which the solution of the given initial value problem is guaranteed to exist: ???????? ′ + tan(????) ???? = sin(????) , ????(????) = 6.
Which line is parallel to line r?
If a ball is dropped near the surface of the earth, then the distance it falls is directly proportional to the square of the time it has fallen. A ball is dropped over the edge of a vertical cliff and falls 39.2 meters in two seconds. Determine the distance (in meters) the ball would have dropped in 3.5 seconds.
Using fermat's little theorem, find the least positive residue of $2^{1000000}$ modulo 17.
By applying Fermat's Little Theorem, we can find the least positive residue of [tex]2^{1000000[/tex] modulo 17. Using the repeated squaring method, we calculate that [tex]2^{16[/tex] is congruent to 1 modulo 17. Therefore, [tex]2^{1000000[/tex] is also congruent to 1 modulo 17.
Fermat's Little Theorem states that if p is a prime number and a is any positive integer that is not divisible by p, then a raised to the power of p-1 is congruent to 1 modulo p. In this case, we need to find the least positive residue of 2 raised to the power of 1000000 modulo 17.
First, we need to find the value of 2 raised to the power of 16 modulo 17, since 17 is a prime number. Using the repeated squaring method, we can calculate:
[tex]2^2[/tex] = 4 (mod 17)
[tex]2^4[/tex] = [tex](2^2)^2[/tex] = [tex]4^2[/tex] = 16 (mod 17)
[tex]2^8[/tex] = [tex](2^4)^2[/tex] = [tex]16^2[/tex] = 1 (mod 17)
[tex]2^{16[/tex] = [tex](2^8)^2[/tex] = [tex]1^2[/tex] = 1 (mod 17)
Now, we can find the value of [tex]2^{1000000[/tex] modulo 17 using the fact that [tex]2^{16[/tex] is congruent to 1 modulo 17:
[tex]2^{1000000[/tex] = [tex](2^{16})^{62500[/tex] x [tex]2^0[/tex] = [tex]1^{62500[/tex] x 1 = 1 (mod 17)
Therefore, the least positive residue of [tex]2^{1000000[/tex] modulo 17 is 1.