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nikklg [1K]
3 years ago
12

Gregory used a calculator to find the equation h = 1.32t + 1.56, where t is the number of days after the grass was trimmed and h

is its height measured in centimeters. Predict the height of the grass after 4 days.
F 1.85 cm
G 6.11 cm
H 6.35 cm
J 6.84 cm
Mathematics
1 answer:
hram777 [196]3 years ago
8 0

Answer:

J

Step-by-step explanation:

h = 1.32t + 1.56

Replace t with 4 (days):

h = 1.32(4) + 1.56

Multiply:

h=5.28 + 1.56

Add:

h=6.84

The correct choice is J

<em>*You don't have to, but I am currently trying to reach the next level, and all I need is some more brainliest answers. If you think my answer was brainly enough, you can make my answer the brainliest, but no pressure. I just help people for fun! :) Thank you, have a great day!*</em>

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sweet-ann [11.9K]

Answer:

1.8

Step-by-step explanation:

If we know Xin ordered 5 identical pieces of bread, this must mean that each one weighs the same

By dividing the 9 pounds with 5 (the amount of bread) we will get 1.8

So each bread is 1.8

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5 0
3 years ago
(-3) + (+1) =<br> steps of solving
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Answer:

-2

Step-by-step explanation:

Opening brackets,

-3+1=-2

3 0
3 years ago
Read 2 more answers
In ΔEFG, the measure of ∠G=90°, the measure of ∠E=78°, and GE = 33 feet. Find the length of FG to the nearest tenth of a foot.
ANTONII [103]

Answer:

I just did this on Delta Math lol:

Answer: 155.3 ft

Step-by-step explanation:

7 0
3 years ago
A simple random sample of 90 is drawn from a normally distributed population, and the mean is found to be 138, with a standard d
bagirrra123 [75]

The 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C:  [130.10, 143.90]

<h3>
How to find the confidence interval for population mean from large samples (sample size > 30)?</h3>

Suppose that we have:

  • Sample size n > 30
  • Sample mean = \overline{x}
  • Sample standard deviation = s
  • Population standard deviation = \sigma
  • Level of significance = \alpha

Then the confidence interval is obtained as

  • Case 1: Population standard deviation is known

\overline{x} \pm Z_{\alpha /2}\dfrac{\sigma}{\sqrt{n}}

  • Case 2: Population standard deviation is unknown.

\overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}

For this case, we're given that:

  • Sample size n = 90 > 30
  • Sample mean = \overline{x} = 138
  • Sample standard deviation = s = 34
  • Level of significance = \alpha = 100% - confidence = 100% - 90% = 10% = 0.1 (converted percent to decimal).

At this level of significance, the critical value of Z is: Z_{0.1/2} = ±1.645

Thus, we get:

CI = \overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}\\CI = 138 \pm 1.645\times \dfrac{34}{\sqrt{90}}\\\\CI \approx 138 \pm 5.896\\CI \approx [138 - 5.896, 138 + 5.896]\\CI \approx [132.104, 143.896] \approx [130.10, 143.90]

Thus, the 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C:  [130.10, 143.90]

Learn more about confidence interval for population mean from large samples here:

brainly.com/question/13770164

3 0
2 years ago
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