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vladimir1956 [14]

4 years ago

14

Marram spent 3/4 of her allowance at the mall. Of the money spent at the mall1/3 was spent on new earrings. What part of her all

owance did Merrimack spent on earrings.

1 answer:

nikitadnepr [17]4 years ago

7 0

3/8 this is your answer so yahhh hope this heped

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Find the Range and Domain of the Binary relations

sukhopar [10]

Answer:

A = {3,5,6}

B = {1,3}

<u><em>A x B</em></u> = {(3,1),(3,3),(5,1),(5,3),(6,1),(6,3)}

Domain of A x B = (The x figures in ascending order)

=> {3,5,6} (Figures repeating more than once must be written only 1 time)

Range of A x B = (The y figures in ascending order)

=> {1,3}

<u><em>B x A </em></u>= {(1,3),(3,3),(1,5),(3,5),(1,6),(3,6)}

Domain of B x A = {1,3}

Range of A x B = {3,5,6}

8 0

4 years ago

Help ASP. what 3/4 divided by 1/8

iris [78.8K]

Answer:

6

Step-by-step explanation:

3/4 / 1/8

3/4 x 8/1

24/4

6

3 0

3 years ago

Read 2 more answers

Help me please thank you.

jenyasd209 [6]

The correct answer is A: 2.

The slope compares the vertical change (the rise) to the horizontal change (the run) when moving from one fixed point to another along the line.

If you look at your graph, it takes 2 units up and 1 unit to the right in order to get to the next point on the line.

Therefore, the slope of the line is 2/1, or simply 2. It has a positive value because the line is sloping upward.

4 0

3 years ago

Find the first four terms for the arithmetic sequence given a1 = 6 and d = 5

DENIUS [597]

Answer:

6, 11, 16, 21

Step-by-step explanation:

To obtain the first 4 terms add the common difference 5 to the previous term, that is

a₁ = 6

a₂ = a₁ + 5 = 6 + 5 = 11

a₃ = a₂ + 5 = 11 + 5 = 16

a₄ = a₃ + 5 = 16 + 5 = 21

7 0

3 years ago

A random sample of 23 items is drawn from a population whose standard deviation is unknown. The sample mean isx⎯ ⎯ x¯ = 840 and

gayaneshka [121]

Answer:

$(832.156, \ 847.844)$

Step-by-step explanation:

Given data :

Sample standard deviation, s = 15

Sample mean, $\overline x = 840$

n = 23

a). 98% confidence interval

$\overline x \pm t_{(n-1, \alpha /2)}. \frac{s}{\sqrt{n}}$

$$E= t_{( n-1, \alpha/2 )} \frac{s}{\sqrt n}}$

$t_{(n-1 , \alpha/2)} \frac{s}{\sqrt n}$

$t_{(n-1, a\pha/2)}=t_{(22,0.01)} = 2.508$

∴ $E = 2.508 \times \frac{15}{\sqrt{23}}$

$E = 7.844$

So, 98% CI is

$(\overline x - E, \overline x + E)$

$(840-7.844 , \ 840+7.844)$

$(832.156, \ 847.844)$

4 0

3 years ago