What is the value of n? Enter your answer in the box.

How many half-lives of radon-222 have passed in 11.46 days? If 5.2 × 10−8 g of radon-222 remain in a sealed box after 11.46 days

Svetach [21]

Answer:

(i) Approximately 3 half lifes

(ii) $4.21\times 10^{-7}\text{ g}$

Step-by-step explanation:

(i) ∵ The half life of Radon-222 is approximately 3.8 days,

So, the number of half life in 11.46 days = $\frac{11.46}{3.8}$ ≈ 3

(ii) Since, the half life formula is,

$N=N_0 (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

Where,

$N_0$ = initial quantity,

t = number of periods

$t_{\frac{1}{2}}$ = half life of the quantity,

Given,

N = $5.2\times 10^{-8}\text{ g}$

t = 11.46 days,

$t_{\frac{1}{2}} = 3.8\text{ days}$

$\implies 5.2\times 10^{-8}=N_0 (\frac{1}{2})^\frac{11.46}{3.8}$

$\implies N_0=2^{\frac{11.46}{3.8}}\times 5.2\times 10^{-8}\approx 4.21\times 10^{-7}\text{ g}$

7 0

4 years ago

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 400 babies were born, a

Masja [62]

Answer:

(a) 99% confidence interval for the percentage of girls born is [0.804 , 0.896].

(b) Yes, the proportion of girls is significantly different from 0.50.

Step-by-step explanation:

We are given that a clinical trial tests a method designed to increase the probability of conceiving a girl.

In the study 400 babies were born, and 340 of them were girls.

(a) Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of girls born = $\frac{340}{400}$ = 0.85

n = sample of babies = 400

p = population percentage of girls born

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.85-2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } }$ , $0.85+2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } }$ ]

= [0.804 , 0.896]

Therefore, 99% confidence interval for the percentage of girls born is [0.804 , 0.896].

(b) Let p = population proportion of girls born.

So, Null Hypothesis, $H_0$ : p = 0.50 {means that the proportion of girls is equal to 0.50}

Alternate Hypothesis, $H_A$ : p $\neq$ 0.50 {means that the proportion of girls is significantly different from 0.50}

The test statistics that will be used here is One-sample z proportion test statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of girls born = $\frac{340}{400}$ = 0.85

n = sample of babies = 400

So, the test statistics = $\frac{0.85-0.50}{\sqrt{\frac{0.85(1-0.85)}{400} } }$

= 19.604

Now, at 0.01 significance level, the z table gives critical value of 2.3263 for right tailed test. Since our test statistics is way more than the critical value of z as 19.604 > 2.3263, so we have sufficient evidence to reject our null hypothesis due to which we reject our null hypothesis.

Therefore, we conclude that the proportion of girls is significantly different from 0.50.

8 0

3 years ago

Analyze the diagram below and complete the instructions that follow.

Aleonysh [2.5K]

m<BDC = m<ADB = 37

answer
A 37

4 0

3 years ago

Read 2 more answers

What is the scale factor of ABC to DEF? A.6 B.8 C.1/8 D.1/6

lord [1]

Answer: