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4 years ago

Exactly how many days is 72 hours?

Mathematics

2 answers:

harina [27]4 years ago

7 0

2 days 3 nights is about 72 hours

jasenka [17]4 years ago

5 0

72 hours is exactly 3 days...

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goldenfox [79]

1. 7/6

2. 30

hope it helped!

4 0

3 years ago

Read 2 more answers

The heights of some sunflowers are shown below.

deff fn [24]

Answer:

2 sunflowers

Step-by-step explanation:

Number of sunflowers that are 7 feet tall = 4

Number of sunflowers that are 5 1/2 feet tall = 2

4 - 2 = 2

8 0

3 years ago

Read 2 more answers

0.2124 divided by 0.06

Strike441 [17]

Answer:

3.54

Step-by-step explanation:

move the decimal place over two making the numbers 6 and 21.24

6 goes into 21 3 times with a remainder of 3 (carry it down)

6 goes into 32 5 times with a remainder of 2 (carry it down)

6 goes into 24 4 times with no remainders

put the numbers together and don't forget the decimal

that number is 3.54

(I was trying to explain long division in words and boy is it difficult)

5 0

3 years ago

The report "The New Food Fights: U.S. Public Divides Over Food Science" states that younger adults are more likely to see foods

zheka24 [161]

Solution :

$n_1=178, \hat p_1 = 0.48$

$n_2=427, \hat p_2 = 0.38$

$\hat p_1=\frac{x_1}{n_1}$

$x_1=n_1 \hat p_1$

= 178 x 0.48

= 85.44

≈ 85

$x_2=n_2 \hat p_2$

= 427 x 0.38

= 162.26

≈ 162

a). Yes, the sample sizes are large enough to use the large sample confidence interval so as to estimate the difference in the population proportions.

Let $\hat p_1 = 0.48, \ \ \hat p_2 = 0.38, n_1 = 178 \ \ n_2 = 427$

Where the $\text{subscript indicates}$ the age $18-29$ group and the 2 - subscript indicates the age $50-64$ group.

Since $n_1 \hat p_1 = 85.44$

$n_1(1- \hat p_1)=92.56, \ \ n_2\hat p_2 = 162.26, \ n_2(1-\hat p_2) = 264.74$

are all at least 10, the sample sizes are large enough to use the large sample confidence interval.

$P_0=\frac{x_1+x_2}{n_1+n_2}$

$=\frac{85.44+162.26}{178+427}$

= 0.409421

$P_0 = 0.4074$

$Q_0=1-P_0$

= 1 - 0.4094

= 0.5906

b). 90% confidence interval is

$=\left( \hat p_1- \hat p_2 \pm \frac{z \alpha}{2} \times \sqrt{P_0Q_0\left(\frac{1}{n_1}+\frac{1}{n_2}}\right) \right)$

$=\left( 0.48-0.38 \pm \frac{z \times 0.10}{2} \times \sqrt{0.4094 \times 0.5906\left(\frac{1}{178}+\frac{1}{427}}\right) \right)$

$=(0.1 \pm z 0.05 \times 0.04387082)$

$=(0.1 \pm 1.64 \times 0.0439)$

$=(0.1 - 0.071996, 0.1+0.071996)$

$=(0.028004, 0.171996)$

$=(0.0280, 0.1720)$

c). Zero is not included in the confidence interval. Answer is no. The difference in the two population proportion are different from each other.

5 0

3 years ago

In a large class of introductory Statistics students, the professor has each person toss a fair coin 1111 times and calculate t

lutik1710 [3]

Answer:

Given:

Sample size, n = 11

P = 0.5

a) The shape of the histogram will be symmetrical. This is because the probability of getting heads and tails is equal.

b) The histogram is centered at

p = 0.5 (because of equal probability of obtaining heads and tails).

c) How much variability would you expect among these proportions?

Here, we are to find the standard deviation.

Let's use the formula:

$\sigma = \sqrt{\frac{pq}{n}}$

Where

p = 0.5(probability of getting heads)

q = 0.5 (probability of getting tails)

Therefore

$\sigma = \sqrt{\frac{0.5 * 0.5}{11}}$

= 0.0227 ≈ 0.023

The standard deviation is 0.023

d) A normal model should not be use here because the success/failure condition is violated, since each student only flips the coin 11 times, it impossible to obtain both at least 10 heads and at least 10 tails. Here, the sample size is too small.

4 0

4 years ago