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neonofarm [45]
3 years ago
13

HELLPPPPP IDK WHAT TO DOOO

Mathematics
1 answer:
vesna_86 [32]3 years ago
6 0

The formula can be written in the form ...

... T = (original temperature) + x·(change in temperature each minute)

The problem statement gives you the original temperature (21 °C) and tells you the temperature after 12 minutes. You have to use that information to figure out the change in temperature each minute.

The change in temperature in 12 minutes is (75 °C) - (21 °C) = 54 °C. That is the change in 12 minutes, so the change in 1 minute will be 1/12 of that:

... 54/12 = 4.5

Using this value in the equation for T, we have

... T = 21 + 4.5x

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What is the ratio (multiplier) for the following geometric sequence? 2, 0.5, 0.125, .03125...
Firlakuza [10]
The multiplier is (1/4) or .25

5 0
3 years ago
Please help asap 25 pts
Kazeer [188]
s=12\sqrt{4t}+10

for\ x=0\to y=12\sqrt{4\cdot0}+10=12\sqrt0+10=10\\for\ x=1\to y=12\sqrt{4\cdot1}+10=12\sqrt4+10=12\cdot2+10=34\\for\ x=4\to y=12\sqrt{4\cdot4}+10=12\sqrt{16}+10=12\cdot4+10=58\\for\ x=6\to y=12\sqrt{4\cdot6}+10=12\cdot2\sqrt6+10=24\sqrt6+10\approx69\\for\ x=10\to y=12\sqrt{4\cdot10}+10=12\cdot2\sqrt{10}+10=24\sqrt{10}+10\approx86\\for\ x=12\to y=12\sqrt{4\cdot12}+10=12\cdot4\sqrt3+10=48\sqrt3+10\approx93

\underline{x|\ 0|\ 1\ |\ 4|\ 6\ |10|12|}\\y|10|34|58|69|86|93|

\text{The graph in attachment}

\boxed{Answer:\ about\ \$93,000}

3 0
4 years ago
A trucking firm has a large inventory of spare parts that have been in storage for a long time. It knowsthat some proportion of
Debora [2.8K]

Answer:

0.1546\leq \widehat{p}\leq 0.3513

Step-by-step explanation:

The firm tests 75 parts, and finds that 0.25 of them are notusable

n = 75

x = 0.25 \times 75 = 18.75≈19

\widehat{p}=\frac{x}{n}

\widehat{p}=\frac{19}{75}

\widehat{p}=0.253

Confidence level = 95%

So, Z_\alpha at 95% = 1.96

Formula of confidence interval of one sample proportion:

=\widehat{p}-Z_\alpha \sqrt{\frac{\widehat{p}(1-\widehat{p}}{n}}\leq \widehat{p}\leq \widehat{p}+Z_\alpha \sqrt{\frac{\widehat{p}(1-\widehat{p}}{n}}

=0.253-(1.96)\sqrt{\frac{0.253(1-0.253)}{75}}\leq \widehat{p}\leq0.253+(1.96)\sqrt{\frac{0.253(1-0.253}{75}}

Confidence interval =0.1546\leq \widehat{p}\leq 0.3513

7 0
4 years ago
Christine purchased a prepaid phone card for $20. Long distance calls cost 18 cents a minute using this card. Christine used her
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Answer:

4 minutes and 32 seconds

Step-by-step explanation:

card $20.00-money left over $15.68            $20.00-$15.68  4.32

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3 years ago
. Select all the properties of a rectangle.
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Answer:

A, C, D

Step-by-step explanation:

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3 years ago
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