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WINSTONCH [101]

3 years ago

13

Help me with a. and b.need explaination and by today

1 answer:

evablogger [386]3 years ago

4 0

A
Convert the temperature 16°F to Celsius:
16 = 1.8С + 32
16 - 32 = 1.8C
-16 = 1.8C
С = -16/1.8 ≈ -9
Answer: 16°F ≈ -9°C

b
Convert the temperature -11°C to °F:
F = 1.8(-11) + 32
F = -19.8 + 32
F = 12.2

Find the difference:
difference = 12.2 - (-80) = 12.2 + 80 = 92.2°F

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The equatorial radius of the planet Jupiter is measured 40 times by a process that is practically free of bias. These measuremen

Nina [5.8K]

Answer:

The 93% confidence interval for the equatorial radius of Jupiter is between 71484 km and 71500 km.

Step-by-step explanation:

Sample size of 30 or larger, so we can use the normal distribution to find the confidence interval.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.93}{2} = 0.035$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.035 = 0.965$ , so $z = 1.81$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.81*\frac{28}{\sqrt{40}} = 8$

The lower end of the interval is the sample mean subtracted by M. So it is 71492 - 8 = 71484 km.

The upper end of the interval is the sample mean added to M. So it is 71492 + 8 = 71500 km.

The 93% confidence interval for the equatorial radius of Jupiter is between 71484 km and 71500 km.

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Greg drives 26 miles on Monday and90 miles on tuesday write and expression to match the words

Georgia [21]

26+90=116miles altogether

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I need help writing the first 5 terms of the sequence!

Angelina_Jolie [31]

The first five terms of the sequence are 1, 4, 7, 10, 13.

Solution:

Given data:

$a_{1}=1$

$a_{n}=a_{n-1}+3$

General term of the arithmetic sequence.

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d = 3

$a_{n}=a_{n-1}+3$

Put n = 2 in $a_{n}=a_{n-1}+3$ , we get

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$a_2=4$

Put n = 3 in $a_{n}=a_{n-1}+3$ , we get

$a_{3}=a_2+3$

$a_{3}=4+3$

$a_3=7$

Put n = 4 in $a_{n}=a_{n-1}+3$ , we get

$a_{4}=a_3+3$

$a_{4}=7+3$

$a_4=10$

Put n = 5 in $a_{n}=a_{n-1}+3$ , we get

$a_{5}=a_4+3$

$a_{5}=10+3$

$a_5=13$

The first five terms of the sequence are 1, 4, 7, 10, 13.

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4 years ago

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slava [35]

Answer:

<em>Parallel</em>

Step-by-step explanation:

The dotted lines are parallel because they have the same <u>gradient</u> to each other, this means the two dotted lines will <u>never</u> touch each other.

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Answer:

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Step-by-step explanation:

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