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3 years ago

HELP ASAP PLEASE {will gave thanks and a brainliest}

Mathematics

1 answer:

Luba_88 [7]3 years ago

8 0

Alright here I go:

9.98 * 10^6 in standard form is 9,980,000
7.3 * 10^7 in standard form is 73,000,000
73,000,000 subtracted by 9,980,000 is 63020000.
63020000 in scientific notation is 6.302 * 10^7
Hope I helped!

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Answer:

Step-by-step explanation:

4/15×6/14= 0.1142

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Write a two-column proof. Given: <2 is congruent to <5; Segment AB is congruent to Segment DE Prove: Segment BC is congrue

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Answer:

proof

Step-by-step explanation:

Statements

Reasons

<2 is congruent to <5; Segment AB is congruent to Segment DE

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3 years ago

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Answer:

x>-10

Step-by-step explanation:

(-9x-3) + 6 < 93

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4 0

3 years ago

Intravenous fluid bags are filled by an automated filling machine. Assume that the fill volumes of the bags are independent, nor

kenny6666 [7]

Answer:

a) 0.0167

b) 0

c) 5.948

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 6.16 ounces

Standard Deviation, σ = 0.08 ounces

We are given that the distribution of fill volumes of bags is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) Standard deviation of 23 bags

$\displaystyle\frac{S.D}{\sqrt{23}} = \frac{0.08}{\sqrt{23}} = 0.0167$

b) P( fill volume of 23 bags is below 5.95 ounces)

P(x < 5.95)

$P( x < 5.96) = P( z < \displaystyle\frac{5.95 - 6.16}{0.0167}) = P(z < -12.57)$

$= 1 - P(z < 12.57)$

Calculation the value from standard normal z table, we have,

$P(x < 5.95) = 1 - 1 = 0$

c) P( fill volume of 23 bags is below 6 ounces) = 0.001

P(x < 6) = 0.001

$P( x < 6) = P( z < \displaystyle\frac{6 - \mu}{0.0167})$

Calculation the value from standard normal z table, we have,

$P( z \leq -3.09) = 0.001$

$\displaystyle\frac{6 - \mu}{0.0167} = -3.09\\\\\mu = 6 + (0.0167\times -3.09) = 5.948$

If the mean will be 5.948 then the probability that the average of 23 bags is below 6.1 ounces is 0.001.

7 0

3 years ago