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3 years ago

How do I solve this problem

Mathematics

1 answer:

yuradex [85]3 years ago

5 0

Is that the full
question, is their a picture with it

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What is the square root of 4.84

dybincka [34]

Answer:

2.2. Hope it helps :D

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2 years ago

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Please answer the ixl math problem for 15 points and please fill in the blanks with the correct answer please i beg will mark br

aivan3 [116]

Answer:

y = 5x + 4

Step-by-step explanation:

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3 years ago

WILL MARK BRAINLIEST NEED HELP ASAP :D

nexus9112 [7]

The measure of angle RML is of 130º.

<h3>What is the value of x?</h3>

Angle NML is the sum of angles NMR and RML, hence:

NML = NMR + RML.

The given measures of the angles are:

NML = 156º.

NMR = 26º.

Hence we can solve the equation for RML to find the measure of angle RML, as follows:

NML = NMR + RML

156 = 26 + RML

RML = 130º.

The measure of angle RML is of 130º.

A similar problem, in which the measures of the angles are calculated, is given at brainly.com/question/25716982

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1 year ago

1. Amanda weighs 48 pounds (lbs). The physician ordered a drug that is delivered in the proportion of milligrams (mg) of medicat

allsm [11]

48 pounds/1 * 1 kg/2.2 pounds=<span>21.82 kg
amanda weighs 21.82 kg, so the amount of medication that needs to be administered should be 21.82 mg.</span>

4 0

3 years ago

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65

erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 2.65

$\sigma$ = standard deviation = 0.85

n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

P(X bar <= 3) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z < 2.06) = 0.98030

P(X bar <= 2.65) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3) = 0.98030 - 0.5 = 0.4803 .

8 0

3 years ago