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Travka [436]
3 years ago
11

The temperature in a greenhouse should be 66 degrees or higher. One​ morning, the heater stopped working. The temperature droppe

d 2 degrees before someone fixed the heater. The temperature was still at least 66 degrees when the heater started working again. How can you best describe the temperature in the greenhouse before the heater stopped​ working?
The temperature in the greenhouse was at least (answer) degrees before the heater stopped working.

Please answer ASAP!
Mathematics
1 answer:
erastova [34]3 years ago
8 0

Answer:

The temperature in the greenhouse was at least 68 degrees before the heater stopped working.

Step-by-step explanation:

It was at 68 then it dropped by 2 degrees.

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The answer is 20% since the percentages cancels each other out.
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Find the volume of the following compound shape.<br>​
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<h3>Given:</h3>
  • Cone
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<h3>Volume of the cone:</h3>

v =  \frac{1}{3} \pi {r}^{2} h

v =  \frac{1}{3}  \times \pi \times  {10}^{2}  \times 10

v = 1047.20 \:  {cm}^{3}

<h3>Volume of the cylinder:</h3>

v = \pi {r}^{2} h

v = \pi \times  {10}^{2}  \times 10

v = 3141.59 \:  {cm}^{3}

<h3>Total volume:</h3>

v = 1047.20 + 3141.59

v = 4188.79 \:  {cm}^{3}

<u>Hence</u><u>,</u><u> </u><u>the</u><u> </u><u>volume</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>given</u><u> </u><u>cone</u><u> </u><u>shape</u><u> </u><u>is</u><u> </u><u>4188.7</u><u>9</u><u> </u><u>cubic</u><u> </u><u>centimeters</u><u>.</u>

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2 years ago
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Write the sum as a product of two factors: 12a + 16b + 8
serious [3.7K]
Answer - 20a + 24b because 8 is added to both variables.
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3 years ago
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How can you use rates of change to solve a real-life problem like the excessive emission of CO2 in the air? Answer using correct
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Answer:

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8 0
2 years ago
Joe used a project management software package and has determined the following results for a given project.: Expected completio
statuscvo [17]

Answer:

0.1151 = 11.51% probability of completing the project over 20 days.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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.

Expected completion time of the project = 22 days.

Variance of project completion time = 2.77

This means that \mu = 22, \sigma = \sqrt{2.77}

What is the probability of completing the project over 20 days?

This is the p-value of Z when X = 20, so:

Z = \frac{X - \mu}{\sigma}

Z = \frac{20 - 22}{\sqrt{2.77}}

Z = -1.2

Z = -1.2 has a p-value of 0.1151.

0.1151 = 11.51% probability of completing the project over 20 days.

4 0
3 years ago
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