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3 years ago

I rly need some help please

Mathematics

1 answer:

kicyunya [14]3 years ago

6 0

Answer:

B) 12.5%

Step-by-step explanation:

1. Find amount spent on clothes in July

$2800 * 8% = j

j = $224

2. Find amount spent of clothes in August

$224 + $126 = a

a = $350

3. Find the amount's percentage of the total

$350/$2800 = x

x = 0.125 = 12.5%

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Solve for x (10 - 5 2/3) X (-2) + x = 8 2/3

Ede4ka [16]

(10 - 5 2/3) * -2 + x = 8 2/3
(9 3/3 - 5 2/3) * -2 + x = 8 2/3
4 1/3 * -2 + x = 26/3
13/3 * -2 + x = 26/3
-26/3 + x = 26/3
x = 26/3 + 26/3
x = 52/3 or 17 1/3 <==

6 0

3 years ago

Please help.i am really struggling

alina1380 [7]

Answer:

3,-3) becomes ; (3 + 5 , -3-12) ; (8,-15)

(7,-10) becomes;( 7 + 5, -10-12) ; (12,-22)

(13,-14) becomes (7 + 13, -14-10) ; (20,-24)

Step-by-step explanation:

What we have to do here is to add 5 to the x-axis value and subtract 12 from the y-axis value

(3,-3) becomes ; (3 + 5 , -3-12) ; (8,-15)

(7,-10) becomes;( 7 + 5, -10-12) ; (12,-22)

(13,-14) becomes (7 + 13, -14-10) ; (20,-24)

3 0

3 years ago

I WILL GIVE BRAINLIEST TO CORRECT ANSWER -x<-x+7(x-2)

BlackZzzverrR [31]

Answer:

x>2

Step-by-step explanation:

0<7x-14

-7x<-14

x>2

hope that's right

8 0

3 years ago

Read 2 more answers

Solve this question for brainlest

g100num [7]

I don’t see no question to answer.

5 0

3 years ago

Elwin Osbourne, CIO at GFS, Inc., is studying employee use of GFS e-mail for non-business communications. A random sample of 200

ladessa [460]

Answer:

The 90% confidence interval for the population proportion is

(0.10872, 0.19128)

Step-by-step explanation:

Explanation:-

Step(i):-

Given data A random sample of 200 e-mail messages was selected. Thirty of the messages were not business related

Given random sample size 'n' = 200

Given Thirty of the messages were not business related

let 'x' = 30

Probability of the messages were not business related or proportion

$p = \frac{x}{n} = \frac{30}{200} = 0.15$

Step(ii):-

The 90% confidence interval for the population proportion is

$(p - Z_{0.10} \sqrt{\frac{p(1-p)}{n} } , p + Z_{0.10} \sqrt{\frac{p(1-p)}{n} } )$

Level of significance ∝ = 0.90 or 0.10

The critical value Z₀.₁₀ = 1.645

The 90% confidence interval for the population proportion is

$( 0.15-1.645 \sqrt{\frac{0.15(1-0.15)}{200} } , 0.15 +1.645 \sqrt{\frac{0.15(1-0.15)}{200} } )$

on calculation, we get

(0.15 - 0.04128 , (0.15 + 0.04128)

(0.10872, 0.19128)

Conclusion:-

The 90% confidence interval for the population proportion is

(0.10872, 0.19128)

5 0

3 years ago