All categories

3 years ago

Compare the fractions. 4/7 1/6. a. > b. = c.

Mathematics

2 answers:

mixas84 [53]3 years ago

6 0

Answer:

Step-by-step explanation:

Let's do it substracting:

$\frac{4}{7} - \frac{1}{6} = \frac{24-7}{42} = \frac{17}{42} > 0$

Then,

$\frac{4}{7} - \frac{1}{6} > 0$

$\frac{4}{7} > \frac{1}{6}$ .

BabaBlast [244]3 years ago

4 0

To start with, write both fractions with same denominator, i.e LCM of 6 and 7
Thus,
4/7=24/42

1/6=7/42
This means 4/7 is greater than 1/6
The answer is thus a

You might be interested in

Which answer choice gives the correct explanation for this problem? 40 ÷ 4

Papessa [141]

I divide 40 into 4 even groups

6 0

3 years ago

Points A(7, -5), B(-1, -2), and C(-4, 6) form triangle ABC on a coordinate plane. According to its sides, triangle ABC is a(n) _

prohojiy [21]

Isosceles Triangle, has 2 equal sides and angles.

7 0

3 years ago

During the 7th examination of the Offspring cohort in the Framingham Heart Study, there were 1219 participants being treated for

AlexFokin [52]

Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

The sample proportion is : $\hat p$ = x/n = 1219/3532 = 0.345

Firstly, the pivotal quantity for 95% confidence interval for the proportion of the population 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 = 0.345

n = sample of participants = 3532

p = population proportion

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p= [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.345-1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ , $0.345+1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ ]