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3 years ago

8

A scientist has 350 mg of a radioactive material. The material will cut in half every month. After how many months will there be

.68mg left?

1 answer:

Zanzabum3 years ago

6 0

The answer would be about 9 months.

You might be interested in

Could you please help me? -0.7-(-2.5)=

Reika [66]

-0.7 - (-2.5) =

-0.7 + 2.5 =

2.5 - 0.7 = 1.8

1.8 is your answer

hope this helps

3 0

3 years ago

Find coordinates of point c

BaLLatris [955]

Ans n solution can refer in the picture

4 0

3 years ago

How do you solve these? Please explain steps as well. Thanks.

posledela

$7x-3=2x-33 \\
7x-2x=-33+3\\
5x=-30\\
x=-6\\\\
3(x-2)=2(2x-1)\\
3x-6=4x-2\\
4x-3x=-6+2\\
x=-4
$

3 0

3 years ago

Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is μd =

swat32

Answer:

Test statistic, $t_{s} = -0.603$ (to 3 dp)

Step-by-step explanation:

Deviation, d = x -y

Sample mean for the deviation

$\bar{d} = \frac{\sum x-y}{n}$

$\bar{d} = \frac{(28-6) + (31-27)+(20-26)+(25-25)+(28-29)+(27-32)+(33-33)+(35-34)}{8} \\\bar{d} = -0.625$

Standard deviation: $SD = \sqrt{\frac{\sum d^{2} - n \bar{d}^2}{n-1} }$

$\sum d^{2} = (28-26)^2 + (31-27)^2 +(20-26)^2 +(25-25)^2 +(28-29)^2 +(27-32)^2 +(33-33)^2 +(35-34)^2\\\sum d^{2} = 63$

$SD = \sqrt{\frac{63 - 8 * (-0.625)^2}{8-1} }$

SD =2.93

Under the null hypothesis, the formula for the test statistics will be given by:

$t_{s} = \frac{ \bar{d}}{s_{d}/\sqrt{n} } \\t_{s} = \frac{- 0.625}{2.93/\sqrt{8} }$

$t_{s} = -0.6033$

6 0

3 years ago

Use the Law of Cosines to find the missing angle. In triangle FGH, HF = 16 ft, FG = 23 ft, and m∠F = 52°. Find the measure of f.

ivolga24 [154]

F = 18 ft.

The law of cosines states
c² = a² + b² - 2ab cos C

Using our information, we have
c² = 23² + 16² - 2(23)(16)cos 52
c² = 529 + 256 - 736cos 52
c² = 785 - 736cos 52
c² = 331.8732

Taking the square root of both sides, we have
c = √331.8732 = 18.22 ≈ 18

5 0

3 years ago