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3 years ago

PLEASE HELP IM GIVING OUT 15 FREE POINTS PLUS BRAINLIEST FOR CORRECT ANSWER !!!

Mathematics

2 answers:

Lisa [10]3 years ago

8 0

Answer:

Jacob's rate is two more than one-third of Marc's rate.

Step-by-step explanation:

Let's rewrite the expression as

y = 2 + ⅓r

Now convert it to a word expression

y = 2 + ⅓ r

↑ ↑ ↑ ↑ ↑ ↑

y is two (more than) (one-third) r.

Replace y with "Jacob's rate" and r with "Marc's rate", and the expression becomes

Jacob's rate is two more than one-third Marc's rate.

s344n2d4d5 [400]3 years ago

7 0

Answer:

The answer to your question is: third option is correct.

Step-by-step explanation:

$\frac{1}{3} r + 2$

The correct answer is: Jacob's rate is two more than one-third Marc's rate.

First option is: 3r + 2

Second option is: 2 + 3r

Fourth option: 1/3 r - 2

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igor_vitrenko [27]

The answer is 50
32 + 1.8(10) = 50
(1.8*10=18)

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3 years ago

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The radioactive substance polonium-210, with a half-life of 138.39 days, is present in tobacco. Suppose that the soot in an ex-s

Ksju [112]

Answer:

16.07 milligrams

Step-by-step explanation:

From the question,

Using the formula of radioactive decay,

R/R' = 2ᵃ/ᵇ..................... Equation 1

Where R = Original mass of the polonium-210, R' = Final mass of polonium-210 after decaying, a = Total disintegration time, b = half life of polonium-210.

Make R' the subject of the equation

R' = R/(2ᵃ/ᵇ)................ Equation 2

Given: R = 100 milligrams, a = 1 years = 365 days, b = 138.39 days

Substitute these values into equation 2

R' = 100/(2³⁶⁵⁰⁰/¹³⁸³⁹)

R' = 100/(6.222)

R' = 16.07 milligrams of plonium-210

3 0

3 years ago

The coordinates of the vertices of △ABC are A(−4, 6) , B(−2, 2) , and C(−6, 2) .

marta [7]

translation of 6 units to the right followed by a reflection in the line x=2.

4 0

3 years ago

Read 2 more answers

Carl has already earned $60 mowing lawns. He earns $15 per lawn. He needs at least $115 for a new string trimmer. Write and solv

nata0808 [166]

Answer:

Step-by-step explanation:

60+15w≥115

-60 -60

15w≥55

w≥4

60+15(4)≥115

60+60≥115

5 0

2 years ago

The data below are the frequency of cremation burials found in 17 archaeological sites. a. Obtain the mean, median, and mode of

Sergeeva-Olga [200]

Answer:

(a) The mean of the data is 275.

(b) The median of the data is 85.

(c) The mode of the data is 45.

(d) The measure of center that works best here will be median.

Step-by-step explanation:

We are given below the frequency of cremation burials found in 17 archaeological sites. Arranging those in <u>ascending order</u> we get;

28, 31, 32, 45, 45, 47, 59, 67, 85, 86, 143, 256, 272, 390, 424, 524, 2141.

(a) The formula for calculating mean for the above data is given by;

Mean, $\bar X$ = $\frac{\sum X}{n}$

= $\frac{28+ 31+ 32+ 45+ 45+ 47+ 59+ 67+ 85+ 86+ 143+ 256+ 272+ 390+ 424+ 524+ 2141}{17}$

= $\frac{4675}{17}$ = 275

So, the mean of the data is 275.

(b) Since, the number of observations (n) in our data is odd (i.e. n = 17) , so the formula for calculating median is given by;

Median = $(\frac{n+1}{2} )^{th} \text{ obs.}$

= $(\frac{17+1}{2} )^{th} \text{ obs.}$

= $(\frac{18}{2} )^{th} \text{ obs.}$

= $9^{th} \text{ obs.}$ = 85

So, the median of the data is 85.

(c) <u>Mode</u> is that value in our data which appears maximum number of times in our data.

So, after observing our data we can see that only number 45 is appearing maximum number of times (2 times) and all other numbers are appearing only once.

So, the mode of the data is 45.

(d) The measure of center that works best here will be median because there are outliers in our data (means extreme values) and mean gets affected by the outliers.

So, the best measure would be median as it represents the middle most value of our data.

4 0

3 years ago