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puteri [66]
3 years ago
11

How do you answer a, b and c, answers and also how you worked it out

Mathematics
1 answer:
Kruka [31]3 years ago
4 0

We can either convert to standard form first or convert to a common multiplier, kinda like a common denominator.  The latter makes more sense if they wanted the result in scientific notation, but let's do it that way anyway.

a)

4.5 × 10⁴ + 3.8  × 10³ = 45 × 10³ + 3.8  × 10³ = 48.8× 10³ = 48,800

Answer:  48,800

b)

4.5 × 10⁴ - 3.8  × 10³ = 45 × 10³ - 3.8  × 10³ = 41.2× 10³ = 41,200

Answer:  41,200

c)

7.2 × 10⁻³ + 6.3  × 10⁻² = 7.2 × 10⁻³ + 63  × 10⁻³ = 70.2 × 10⁻³

       = 7.02 × 10⁻² = 0.0702

Answer: 0.0702

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Evaluate the expression 5x-3x+5 for each value of x
kirza4 [7]

Answer:

2x+5

Step-by-step explanation:

Hope I helped :)

4 0
2 years ago
Anyone know this answer?
damaskus [11]
La of sine:

sinC/c = sinB/b==> sin  37°/8 = sin B/12 ==> sin B = 0.903


arcsinB or sin⁻¹ B = 64.5°, & sin (B°) = sin (180° - B°), then

sin(64.5) = sin(180°-64.5°) ==> B = 64.5° or 115.5°
6 0
3 years ago
What’s greater 14/7 or 2/4
aleksandr82 [10.1K]
I think 2/4 but I'm not 100% sure sry
8 0
3 years ago
A nurse opens a case that has a total of 160 ounces of med. If each vial in the case holds 1 1/3oz of medication how many vials
uysha [10]

Answer:

120 Vials

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160/1and1/3

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4 0
1 year ago
Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave
Colt1911 [192]
Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \  \frac{a-\mu}{\sigma}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\  \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\  \\ =1-P(z\ \textless \ -0.5410) \\  \\ =1-0.29426=\bold{0.7057}



Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span>P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}
7 0
3 years ago
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