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Yuliya22 [10]
2 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]2 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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Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):
Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, \mu_X represent the population mean for Brand B and let \mu_Y represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

        P.Q. = \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } ~ t_n__1+n_2-2

where, Xbar = Sample mean for Brand B data = 36.9

            Ybar = Sample mean for Brand A data = 32.9

              n_1  = Sample size for Brand B data = 13

              n_2 = Sample size for Brand A data = 9

              s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2}  }{n_1+n_2-2} } = \sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} } = 3.013

Here, s^{2}_X and s^{2} _Y are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < t_2_0 < 2.528) = 0.98

P(-2.528 < \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } < 2.528) = 0.98

P(-2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (Xbar -Ybar) -(\mu_X-\mu_Y) < 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

P( (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (\mu_X-\mu_Y) < (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} , (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ]

[ (36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} , (36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

                     

4 0
2 years ago
Evaluate each expression for the given values of the variables. SHOW ALL WORK
Reika [66]

Answer:

3(2c+d)−d

Distribute:

=(3)(2c)+(3)(d)+−d

=6c+3d+−d

Combine Like Terms:

=6c+3d+−d

=(6c)+(3d+−d)

=6c+2d

=6c+2d

Step-by-step explanation:

3 0
3 years ago
a fishing tackle box is 13 inches long 6 inches wide and 2 1/2 inches high. what is the volume of the tackle box
grandymaker [24]

Answer:

195 cubic inches

Step-by-step explanation:

A fishing tackle box is 13 inches long, 6 inches wide, and inches high. What is the volume of the tackle box? SOLUTION: The volume of the tackle box is 195 cubic inches.

7 0
2 years ago
Help please thank you
valentinak56 [21]

Answer:

Step-by-step explanation: letter B is the right answer

5 0
3 years ago
Read 2 more answers
Calculate the pay for the following day of a<br> weekly time card given a wage of $12.50/hr.
Elenna [48]

The one day pay is $106.25 rounded to the nearest hundredth.

<u>Step-by-step explanation:</u>

<u>From the table shown :</u>

  • The timing shown in the morning is from 8:00 to 12:15
  • The number of hours worked in the morning = 4 hours 15 minutes.

It is given that, the pay is $12.5 per hour.

Therefore, the pay earned in the morning = No.of hours × pay per hour.

⇒ 4 hours × 12.5 = $50

⇒ (15 mins / 60 mins) × 12.5 = $3.125

⇒ 50+3.125

⇒ 53.125

  • The timing shown in the afternoon is from 8:00 to 12:15
  • The number of hours worked in the morning = 4 hours 15 minutes.

Therefore, the pay earned in the afternoon = No.of hours × pay per hour.

⇒  4 hours × 12.5 = $50

⇒ (15 mins / 60 mins) × 12.5 = $3.125

⇒ 50+3.125

⇒ 53.125

The pay for 1 day = pay earned in the morning section + pay earned in the afternoon section.

⇒ 53.125 + 53.125

⇒ 106.25

∴ The one day pay is $106.25 rounded to the nearest hundredth.

7 0
2 years ago
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