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3 years ago

What is 0.7, 0.755, 5/8 from least to greatest?

1 answer:

6 0

I like to start by turning all my numbers into the same form. You can go either way but I am going to turn the decimals into fractions. so 0.7 is the same as 7/10.

0.755 is like 755/1000 which can be reduced to 151/200 (if you don't understand the convertions let me know i will explaine more.)

Next give all the fractions a common denominator. both 8 and 10 go into 200 so we can use that. 151/200 already has a denominator of 200 so that one is set.
for 7/10 well, 10 time 20 is 200 so multiply both the numerator and the denominator by 20 so you get 140/200
the last one is 5/8 8 times 25 is 200 so multiply both numbers by 25 to get 125/200

the fractionos we have are 125/200 151/200 and 140/200. Now because they all have the same denominator the one with the lowest numerator is the least number and the one with the highest number is the greatest number.

125/200 140/200 151/200 now just replace the original number with the order they are placed in.
125/200 was 5/8
140/200 was 0.7
151/200 was 0.755
so the final answer is
5/8, 0.7, 0.755

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Write 60% as a

Setler79 [48]

Answer:

The answer is three fiths

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3 years ago

In a random sample of 400 residents of Boston, 320 residents indicated that they voted for Obama in the last presidential electi

coldgirl [10]

Answer:

C.I = 0.7608 ≤ p ≤ 0.8392

Step-by-step explanation:

Given that:

Let consider a random sample n = 400 candidates where 320 residents indicated that they voted for Obama

probability $\hat p = \dfrac{320}{400}$

= 0.8

Level of significance ∝ = 100 -95%

= 5%

= 0.05

The objective is to develop a 95% confidence interval estimate for the proportion of all Boston residents who voted for Obama.

The confidence internal can be computed as:

$=\hat p \pm Z_{\alpha/2} \sqrt{\dfrac{ p(1-p)}{n } }$

where;

$Z_{0.05/2}$ = $Z_{0.025}$ = 1.960

SO;

$=0.8 \pm 1.960 \sqrt{\dfrac{ 0.8(1-0.8)}{400 } }$

$=0.8 \pm 1.960 \sqrt{\dfrac{ 0.8(0.2)}{400 } }$

$=0.8 \pm 1.960 \sqrt{\dfrac{ 0.16}{400 } }$

$=0.8 \pm 1.960 \sqrt{4 \times 10^{-4}}$

$=0.8 \pm 1.960 \times 0.02}$

$=0.8 \pm 0.0392$

= 0.8 - 0.0392 OR 0.8 + 0.0392

= 0.7608 OR 0.8392

Thus; C.I = 0.7608 ≤ p ≤ 0.8392

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3 years ago

What is the length of BC ?

MrRa [10]

Looking at this triangle, we can see that is a form of a 45, 45, 90 triangle
This would mean that 2x-24=x-2
Solve: 2x-24=x-2
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3 years ago

Student performance across discussion sections: A professor who teaches a large introductory statistics class (197 students) wit

adell [148]

Answer:

The test is not significant at 5% level of significance, hence we conclude that there's no variation among the discussion sections.

Step-by-step explanation:

Assumptions:

1. The sampling from the different discussion sections was independent and random.

2. The populations are normal with means and constant variance

$H_0:$ There's no variation among the discussion sections

$H_1:$ There's variation among the discussion sections

$\alpha =0.05$

Df Sum Sq Mean sq F value Pr(>F)

Section 7 525.01 75 1.87 0.99986

Residuals 189 7584.11 40.13

Test Statistic = $F= \frac{75}{40.13} =1.87$

$Pr(>F) = 0.99986$

Since our p-value is greater than our level of significance (0.05), we do not reject the null hypothesis and conclude that there's no significant variation among the eight discussion sections.

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3 years ago

As a rule of thumb, the sampling distribution of the sample proportion can be approximated by a normal probability distribution

Gennadij [26K]

As a rule of thumb, the sampling distribution of the sample proportion can be approximated by a normal probability distribution whenever the sample size is large.

<h3>What is the Central limit theorem?</h3>

The Central limit theorem says that the normal probability distribution is used to approximate the sampling distribution of the sample proportions and sample means whenever the sample size is large.
Approximation of the distribution occurs when the sample size is greater than or equal to 30 and n(1 - p) ≥ 5.

Thus, as a rule of thumb, the sampling distribution of the sample proportions can be approximated by a normal probability distribution when the sample size is large and each element is selected independently from the same population.

Learn more about the central limit theorem here:

brainly.com/question/13652429

#SPJ4

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2 years ago