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2 years ago

I'm doing Standard Index Form(s) What is 12 000 000 000 000 in standard form?

Mathematics

1 answer:

Gala2k [10]2 years ago

8 0

Answer:

$12000000000000 = 12 \times {10}^{12}$

<h2>HOPE IT HELPS YOU ✌️</h2>

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Jenny served one scoop of ice cream to her friends using a 3/4 cup scooper. She

Pepsi [2]

Answer:

75?

Step-by-step explanation:

1/4 = 25

25 x 3 =75

----

3/4=75

4 0

2 years ago

Help please!!! I really need it

jenyasd209 [6]

Answer:

140

Step-by-step explanation:

2000 divided by 7% will give you 140 hope it helps

6 0

3 years ago

Write three ratios that are equal to 60/180

Fudgin [204]

1/3 or 3/9 or 9/27
I hope this helps

4 0

2 years ago

Jar contains 8 green, 4 blue, 10 red, and 2 yellow skittles. A skittle is randomly drawn replaced then another is drawn. Find ea

Brrunno [24]

Answer:

green= 33.33%

blue= 16.7%

red=41.6%

yellow=8.33%

Step-by-step explanation:

Total 24 skittles

Divide them

green

8/24

blue

4/24

red

10/24

yellow

2/24

7 0

3 years ago

Read 2 more answers

The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co

Alona [7]

Answer:

95% Confidence interval for the variance:

$3.6511\leq \sigma^2\leq 34.5972$

95% Confidence interval for the standard deviation:

$1.9108\leq \sigma \leq 5.8819$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

$s^2=2.89^2=8.3521$

The confidence interval for the variance is:

$\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}$

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

$\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128$

Then, the confidence interval can be calculated as:

$\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972$

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

$\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819$

6 0

3 years ago