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3 years ago

1/7+2/3= plz help ive been stuck on the question

Mathematics

2 answers:

kotykmax [81]3 years ago

5 0

Answer:

17/21

Step-by-step explanation:

1/7 + 2/3

1/7 = 3/21

2/3 = 14/21

then:

1/7 + 2/3 = 3/21 + 14/21

= (3+14)/21

= 17/21

Romashka [77]3 years ago

3 0

Answer:

1/7+2/3

3+14/21

17/21

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siniylev [52]

Always true because addition and subtraction is commutative
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3 0

2 years ago

Robert went to the grocery store and purchased cans of soup and frozen dinners. each can of soup has 300 mg of sodium and each f

Paraphin [41]

There are 9 cans of soup and 5 cans of frozen dinner.

The variables used for soup and frozen dinner are x and y respectively.

Step-by-step explanation:

Let,

Soup = x

Frozen Dinner = y

According to given statement of sodium;

300x + 500y = 5200 eqn 1

And,

x + y = 14 eqn 2

$Multiplying\ equation\ 2\ with\ 300\\300(x+y=14)\\300x+300y=4200\ Eqn 3\\$

Subtracting Eqn 3 from Eqn 1;

$(300x+500y)-(300x+300y)=5200-4200\\300x+500y-300x-300y=1000\\200y=1000\\y=\frac{1000}{200} \\y=5$

Putting value of y in Eqn 1;

$300x+500(5)=5200\\300x+2500=5200\\\\Subtracting\ 2500\ from\ both\ sides\ \\300x+2500-2500=5200-2500\\300x=2700\\x=\frac{2700}{300} \\x=9$

There are 9 cans of soup and 5 cans of frozen dinner.

The variables used for soup and frozen dinner are x and y respectively.

Keywords: Variables, linear equations, subtraction.

Learn more about linear equations at;

#LearnwithBrainly

5 0

2 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

Viefleur [7K]

Answer:

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

Suppose a sample of 1537 tenth graders is drawn. Of the students sampled, 1184 read above the eighth grade level. So 1537 - 1184 = 353 read at or below this level. Then

$n = 1537, \pi = \frac{353}{1537} = 0.2297$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 - 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2087$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 + 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2507$

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

7 0

3 years ago

What is the volume of a rectangular prism with a length of 2.5 cm, a width of 3.1 cm and a height of 1.2 cm?

Svetradugi [14.3K]

The answer is D- 9.3cm^3

7 0

2 years ago

A grocery store sells three different quantities of sugar. A 1-pound bag costs $1.10, a 2-pound bag costs $1.92, and a 3-pound b

Mashcka [7]

$1.10 per pound for the 1-pound bag.
$0.96 per pound for the 2-pound bag.
$0.92 per pound for the 3-pound bag.

As the quantity of sugar increases, the unit cost decreases.

8 0

3 years ago