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3 years ago

After 3 days a sample of radon-222 has decayed to 58% of its original amount.

1 answer:

6 0

The half-life of the radioactive substance can be determined through teh equation ln 2 = k (t 1/2) where k is equal to ln (1/(1-0.58)) / 3. Hence t 1/2 is equal to 2.40 days. To decay to 10% of the original amount, ln (1/0.90) / ( ln2) / 2.40, time is equal to 0.37 days.

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Weights of the vegetables in a field are normally distributed. From a sample Carl Cornfield determines the mean weight of a box

pantera1 [17]

To find the z-score for a weight of 196 oz., use

$z=\frac{x-\mu}{\sigma}=\frac{196-180}{8}=\frac{16}{8}=2$

A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.

1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.

7 0

3 years ago

Cassandra assigns values to some of the measures of triangle ABC. If angle A measures 30°, a = 6, and b = 18, which is true?

Anna [14]

Answer:

The triangle does not exist because sin(A)/a can not be equal to sin(B)/b

Step-by-step explanation:

we know that

step 1

Find the measure of angle B

Applying the law of sines

$\frac{sin(A)}{a}=\frac{sin(B)}{b}$

we have

$A=30\°$

$sin(30\°)=1/2$

$a=6\ units$

$b=18\ units$

substitute

$\frac{(1/2)}{6}=\frac{sin(B)}{18}$

$\frac{1}{12}=\frac{sin(B)}{18}$

$sin(B)=\frac{18}{12}$

$sin(B)=\frac{3}{2}=1.5$

Remember that the value of sine can not be greater than 1

therefore

The triangle does not exist because sin(A)/a can not be equal to sin(B)/b

8 0

3 years ago

Do the ratios 8 12 and 3 2 form a proportion?

sashaice [31]

Answer:

yea

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

If lines |l, corresponding 2's = Segment RA = segment RB, RS||AB

Debora [2.8K]

Answer:

gang 555s

Step-by-step explanation:

IFYKYK

7 0

2 years ago

Which of the following functions f : {0, 1, 2, 3} ! {0, 1, . . . , 7} are one-to-one?

allsm [11]

Answer:

1. No.

$f(0)=0\\1^2=1; \text{then } f(1)=1\\2^2=4\equiv 4 \text{ mod 8}; \text{then } f(2)=4\\3^2=9\equiv 1 \text{mod 8 }; \text{then } f(3)=1$

Since f(1)=f(3) and $1\neq 3$ then f isn't one-to-one.

2. No

$f(0)=0\\1^3=1\equiv 1\text{ mod 8}; \text{then } f(1)=1\\2^3=8\equiv 0\text{ mod 8}; \text{then } f(2)=0\\3^3=27\equiv 3 \text{ mod 8}; \text{then } f(3)=3$

Since f(0)=f(2) and $0\neq 2$ then f isn't one-to-one.

3. No

$0^3-8=-8\equiv 0\text{ mod 8}; \text{then } f(0)=0\\1^3-8=-7\equiv 1\text{ mod 8}; \text{then } f(1)=1\\2^3-8=0\equiv 0 \text{ mod 8}; \text{then } f(2)=0\\3^3-8=27-8=19\equiv 3 \text{ mod 8}; \text{then } f(3)=3\\$

Since f(0)=f(2) and $0\neq 2$ then f isn't one-to-one.

4. Yes

$0^3+2*0=0; \text{then } f(0)=0\\1^3+2*1=3\equiv 3\text{ mod 8}; \text{then } f(1)=3\\2^3+2*2=8+4=12\equiv 4 \text{ mod 8}; \text{then } f(2)=4\\3^3+2*3=27+6=33\equiv 1\text{ mod 8}; \text{then } f(3)=1$

Since $f(0)\neq f(1)\neq f(2) \neq f(3)$ , then f is one-to-one

5. Since f(1)=f(3) and $1\neq 3$ then, f isn't one-to-one

3 0

3 years ago