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4 years ago

Colin spends

Mathematics

2 answers:

aliina [53]4 years ago

6 0

Answer:

E110

Step-by-step explanation:

well there is E176 (sorry for bad symbol) and then he spends 1/4+1/8 or 3/8 on food so then its 3/8*176 which is 66. 176-66 is 110

HOpe this helps plz hit the crown :D

dlinn [17]4 years ago

4 0

Rent: £176/4 = £44.
Food: £176/8 = £22.
£176-£66=£110

£110 Left.

P.S. Make sure the question says how much he has left per week and not per month!

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5% tax what is it as a fraction and a decimal

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4 years ago

Solve for x: 1 over 3 (2x − 8) = 4. (1 point)

STALIN [3.7K]

You ANSWER : x=10 Let's solve !
Explanation : Let's solve your equation step-by-step.
1
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(
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6 0

4 years ago

Read 2 more answers

2 ( n - 7 ) = 20<br><br> What is n?

Sunny_sXe [5.5K]

Answer:

The answer is n = 17

Step-by-step explanation:

Divide both sides by the numeric factor on the left side, then solve.

Hoped this helped!

brainly, please?

8 0

4 years ago

Having trouble solving this:

aliya0001 [1]

Answer:

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6 0

3 years ago

The following information was obtained from independent random samples taken of two populations. Assume normally distributed pop

Volgvan

Answer:

1. The 95% confidence interval for the difference between means is (-5.34, 11.34).

2. The standard error of (x-bar1)-(x-bar2) is 4.

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=10 has a mean of 45 and a standard deviation of √85=9.2195.

The sample 2, of size n2=12 has a mean of 42 and a standard deviation of √90=9.4868.

The difference between sample means is Md=3.

$M_d=M_1-M_2=45-42=3$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

The critical t-value for a 95% confidence interval is t=2.086.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=2.086 \cdot 4=8.34$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = 3-8.34=-5.34\\\\UL=M_d+t \cdot s_{M_d} = 3+8.34=11.34$

The 95% confidence interval for the difference between means is (-5.34, 11.34).

6 0

4 years ago