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Yuliya22 [10]
3 years ago
12

The ratio of Val's savings to Dan is 2:3 at first. After Dan spent P3000.00, Val's saving became 11/3 more of that Dan's remaini

ng savings. What is the total amount of saving the two had at first
Mathematics
1 answer:
luda_lava [24]3 years ago
8 0

Answer:

P.7500.00

Step-by-step explanation:

We are told that:

The ratio of Val's savings to Dan is 2:3 at first.

Now, Dan spent P3000.00 and Val's saving became 1⅓ more of that Dan's remaining savings.

Let's say the total amount of savings they had at first was x.

Thus;

Val had: 2x/5

Dan had: 3x/5

Now, Dan spent P3000.00.

So amount Dan has left = (3x/5) - 3000

We are told Val's savings is now 1⅓ = 4/3 times that of Dan's remaining savings

Thus; 2x/5 = (4/3)((3x/5) - 3000)

2x/5 = 12x/15 - 4000

Multiply through by 15 to get;

6x = 12x - (3000 × 15)

6x = 12x - 45000

Rearranging, we have;

12x - 6x = 45000

6x = 45000

x = 45000/6

x = P.7500

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6 0
3 years ago
You own an accessories store, and sales last month were $24,000. You had $3,500 in discounts and $975 in returns.
Len [333]

Answer:

The net sales for last month were <u>$19,525</u>.

Step-by-step explanation:

Given:

Last month sales were $24,000.

Discounts is $3,500 and $975 in returns.

Now, to get the net sales for last month.

So, we deduct the discount:

<em>Sales - discounts</em> = \$24,000 - \$3,500 =\$20,500.

Then, we deduct the returns from the remaining amount:

<em>Sales after discounts - returns</em> = \$20,500 - \$975

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Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standar
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We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack

Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack

Where (from tables):

Z_{0.99}=2.33

Finally, the interval at 98% confidence level is:

CI(\mu)=\lbrack28.94,31.06\rbrack

4 0
1 year ago
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We take positive value of sinθ because 0° < θ < 90°
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\sin2\theta=2\sin\theta\cos\theta=2\cdot\dfrac{3}{5}\cdot\dfrac{4}{5}=\dfrac{2\cdot3\cdot4}{5\cdot5}=\dfrac{24}{25}\\\\\\\cos2\theta=\cos^2\theta-\sin^2\theta=\left(\dfrac{4}{5}\right)^2-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}-\dfrac{9}{25}=\dfrac{7}{25}\\\\\\&#10;\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=\dfrac{\frac{24}{25}}{\frac{7}{25}}=\dfrac{24\cdot25}{7\cdot25}=\dfrac{24}{7}=3\dfrac{3}{7}
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Step-by-step explanation:

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