Answer:
The angle of elevation of the sun is 39⁰
Step-by-step explanation:
Given;
height of the tree, h = 96 ft
length of the shadow, L = 120 ft
|
| 96ft
|
|
θ------------------------------------
120ft
Completing this triangle to cut across the top of the tree gives you a right angled triangle with θ as the angle of elevation of the sun.
Apply trig-ratio to determine the angle of elevation of the sun;
tanθ = opposite side / adjacent side
tanθ = 96 / 120
tanθ = 0.8
θ = tan⁻¹(0.8)
θ = 38.7⁰
θ = 39⁰
Therefore, the angle of elevation of the sun is 39⁰
Answer:
66-67 times
Step-by-step explanation:
i would go with 67
you just divide 1.5 by a 100
If the total monthly bill for one child was $750, what was the total number of hours the child spent in after-school care?
A.100
B.90
C.88
D.80
Answer:
D.) 80
Step-by-step explanation:
750 - 350 = 400
400/5 = 80
80 hours
Hopefully this helps you :)
pls mark brainlest ;)
Answer:
D-80
Step-by-step explanation:
750-350=400
400/5=80
Answer: Write an augmented matrix for the system. Then state the dimensions. x+8y−7z=12 5x+9y+5z=15 6z−3y−8x=1
Step-by-step explanation: x+8y-7z=12 5x+9y+5z=15 6z-3y-8x=1. - 18028782.
The system of equations provided can be converted into an augmented matrix as follows: [[1, 8, -7, 12], [5, 9, 5, 15], [-8, -3, 6, 1]]. The dimensions of this matrix are 3x4.
The provided system of equations is:
1. x + 8y - 7z = 12
2. 5x + 9y + 5z = 15
3. -8x - 3y + 6z = 1
We can represent this system as an augmented matrix by aligning the coefficients of the variables and the constants. The augmented matrix is:
[[1, 8, -7, 12], [5, 9, 5, 15], [-8, -3, 6, 1]]
The dimensions of this matrix represents the number of rows and columns it has. In this case, this matrix is a 3x4 matrix because it has 3 rows and 4 columns.
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P' =
Q' =
R' =
Answer:
(x +3)^2 = 0
Step-by-step explanation:
The square is already complete. We can write the equation as a square:
(x +3)^2 = 0
a. If the sample variance is s^2=32 , are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha=.05
b. If the sample variance is s^2=72 , are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha=.05 ?
c. Comparing your answer for parts a and b, how does the variability of the scores in the sample influence the outcome of a hypothesis test?
A hypothesis test was conducted to evaluate the treatment's effect. For both variances, we failed to reject the null hypothesis, so we can't conclude that the treatment had a significant effect. The variability of scores plays a crucial role, as more variability makes it harder to identify a significant effect.
To determine if the treatment has a significant effect, we perform a hypothesis test using the sample mean (M), sample variance (s^2), and population mean (μ). The null hypothesis is that there's no effect from the treatment (μ=M), while the alternative hypothesis is that there is an effect (μ≠M).
a. For sample variance s^2=32, we can use the formula for the t score: t = (M - μ)/(s/√n) = (35 - 40)/(√32/√8) = -2.24. Based on a two-tailed t-distribution table, the critical t values for α=.05 and 7 degrees of freedom (n-1) are approximately -2.365 and 2.365. Our t value (-2.24) lies within this range, so we fail to reject the null hypothesis. We cannot conclude that the treatment has a significant effect.
b. Repeat the same process with sample variance s^2=72. The t value is now (35 - 40)/(√72/√8) = -1.48, again falling within the range of the critical t values. We can't conclude that the treatment has a significant effect.
c. As the variability (s^2) of the sample scores increases, it becomes more difficult to find a significant effect. Higher variability introduces more uncertainty, which can mask actual changes caused by the treatment.
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To evaluate the effect of a treatment using a two-tailed test with alpha = 0.05, we compare the calculated t-value to the critical t-value. The sample variance influences the outcome of the hypothesis test, with a larger variance leading to a wider critical region.
a. To test if the treatment has a significant effect, we will conduct a two-tailed hypothesis test using the t-distribution. The null hypothesis states that the treatment has no effect (μ = 40), while the alternative hypothesis states that the treatment has an effect (μ ≠ 40). With a sample size of 8, degrees of freedom (df) will be n-1 = 7. We will use the t-test formula to calculate the t-value, and compare it to the critical t-value from the t-table with α = 0.05/2 = 0.025. If the calculated t-value falls outside the critical region, we reject the null hypothesis and conclude that the treatment has a significant effect.
b. Similar to part a, we will conduct a two-tailed t-test using the same null and alternative hypotheses. With a sample size of 8, df = n-1 = 7. We will calculate the t-value using the sample mean, population mean, and sample variance. Comparing the calculated t-value to the critical t-value with α = 0.05/2 = 0.025, if the calculated t-value falls outside the critical region, we reject the null hypothesis and conclude that the treatment has a significant effect.
c. The variability of the scores in the sample, as indicated by the sample variance, influences the outcome of the hypothesis test. In both parts a and b, the sample variance is given. A larger sample variance (s^2 = 72 in part b) indicates more variability in the data, meaning the scores in the sample are more spread out. This leads to a larger t-value and a wider critical region. Therefore, it becomes easier to reject the null hypothesis and conclude that the treatment has a significant effect.
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