Find the missing side. 31° Z z = [?] Round to the nearest tenth. Remember: SOHCAHTOA 21​

Answer:

1.50 rand

Step-by-step explanation:

divide 18 by 12

= 1.5

therefore each egg costs 1 rand 50

Answer:

2/3

Step-by-step explanation:

if a dozen of eggs costs 18 rupees, what is the cost of each egg?

to get this we have to divide the cost of 12 eggs to 18 rupees

12/18

2/3 that's all

It would cost 2025p to send one parcel weighing 170g and two parcels weighing 320g each.

In order to find out how much it will cost to send one parcel weighing 170g and two parcels weighing 320g each, we can use the information given to us in the question: it costs 50p to post a parcel that weighs 20g. Using this information, we can create a proportion. We know that 50p is the cost to send a parcel weighing 20g, so if we multiply the weight of the parcel by 2.5, we can find out how much it would cost to send that parcel.

For example, a parcel weighing 50g would cost 125p (50 x 2.5 = 125). To find out how much it would cost to send a parcel weighing 170g, we can use this porpotion: 50p/20g = x/170g

Multiplying both sides by 170g, we get:

x = (50p/20g) * 170gx = 425p

So, it would cost 425p to send a parcel weighing 170g.To find out how much it would cost to send two parcels weighing 320g each, we can use the same proportion:

50p/20g = x/320g

Multiplying both sides by 320g, we get:

x = (50p/20g) * 320gx = 800p

So, it would cost 800p to send two parcels weighing 320g each.

To find out the total cost to send all three parcels, we simply add up the costs:

425p + 800p + 800p = 2025p.

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Answer:

The lower quartile of the fish lengths is 15 inches

The upper quartile of the fish lengths is 30 inches

Step-by-step explanation:

(the cumulative frequency gives how much of the observations/measurements are under a certain value e.g, in our case, for lengths of 40 inches, all 16 of the fish have length in that range (less than or equal to 40) so, the cumulative frequency is 16 and so on)

We need to find the lower and upper quartiles of the fish lengths,

Now, the lower quartile contains 25% (or 1/4) of the values

in our case, since there are 16 fish,

25% of 16 is 4

So, for the lower quartile, the cumulative frequency should be 4.

For which the length is 15 inches,

So, the lower quartile of the fish lengths is 15 inches

The upper quartile contains 75% (or 3/4) of the values,

Now, 3/4 of 16 is 12.

So, for the upper quartile, the cumulative frequency should be 12.

For which the length is 30 inches

So, the upper quartile of the fish lengths is 30 inches

Based on the given information and adjustments, the adjusted bank balance for ABC Company on July 31 would be Br. 5,169.42.

Based on the provided information, let's analyze the various transactions and their impact on the bank reconciliation for ABC Company:

1. The deposit of Br. 410.90 made after banking hours: This deposit does not appear in the bank statement, so it should be added to the bank balance. After including this deposit, the adjusted bank balance would be Br. 5,375.37 (Br. 4,964.47 + Br. 410.90).

2. The erroneously charged check of Br. 79: This check should be deducted from the bank balance as it was charged in error by the bank. After deducting this amount, the adjusted bank balance would be Br. 5,296.37 (Br. 5,375.37 - Br. 79).

3. Outstanding checks: The outstanding checks that haven't been paid by the bank need to be deducted from the adjusted bank balance. The total amount of outstanding checks is Br. 610.75 (Br. 100 + Br. 10.25 + Br. 294.50 + Br. 205). After deducting this amount, the adjusted bank balance would be Br. 4,685.62 (Br. 5,296.37 - Br. 610.75).

4. Incorrectly charged check of Br. 210 as Br. 120: This error made by the bank needs to be corrected. The bank balance should be increased by the difference, which is Br. 90. Therefore, the adjusted bank balance would be Br. 4,775.62 (Br. 4,685.62 + Br. 90).

5. Proceeds from collection of an interest-bearing note receivable: The face amount of the note was Br. 500. Since the note was left with the bank's collection department, the amount collected should be added to the bank balance. Therefore, the adjusted bank balance would be Br. 5,275.62 (Br. 4,775.62 + Br. 500).

6. Interest earned on the average account balance: The interest earned of Br. 24.75 should be added to the bank balance. After including this amount, the adjusted bank balance would be Br. 5,300.37 (Br. 5,275.62 + Br. 24.75).

7. The returned check of Br. 10: This check was recorded in the check register as Br. 100 but was returned for Br. 10. The difference of Br. 90 should be deducted from the bank balance. Therefore, the adjusted bank balance would be Br. 5,210.37 (Br. 5,300.37 - Br. 90).

8. Fee charged by the bank for handling the collection of the note receivable: The fee of Br. 5.00 should be deducted from the bank balance. After deducting this fee, the adjusted bank balance would be Br. 5,205.37 (Br. 5,210.37 - Br. 5.00).

9. Check from customer John charged as Non-sufficient funds (NSF): The check of Br. 50.25 should be deducted from the bank balance. After deducting this amount, the adjusted bank balance would be Br. 5,155.12 (Br. 5,205.37 - Br. 50.25).

10. Service charge by the bank for the month of July: The service charge of Br. 12.70 should be deducted from the bank balance. After deducting this charge, the adjusted bank balance would be Br. 5,142.42 (Br. 5,155.12 - Br. 12.70).

11. Erroneously recorded check of Br. 85 as Br. 58: This error in the cash payment journal needs to be corrected. The bank balance should be increased by the difference, which is Br. 27. Therefore, the adjusted bank balance would be Br. 5,169.42 (Br. 5,142.42 + Br. 27).

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Answer:

constant is - 19

Step-by-step explanation:

the constant is the term in an expression with no variable attached to it.

12r + [tex]\frac{r}{2}[/tex] - 19

the only term without the variable r attached to it is - 19

then the constant term is - 19

Answer:

x+100

Step-by-step explanation:

x^2-10,000/x-100

x^2-10000 is a difference in squares and can be factored.

x^2-10000 = (x+100)(×-100).

(x+100)(x-100)/(x-100)

(x-100) cancel each other.

x+100 remains.

The area of the composite figure is 800m²

The area of a figure is the number of unit squares that cover the surface of a closed figure.

Composite geometric figures are made from two or more geometric figures.

The figure consist of a rectangle , a semi circle and a triangle.

Area of the semicircle = 1/2 πr²

= 1/2 × 3.14 × 10²

= 314/2 = 157 m²

Area of the rectangle = l × w

= 25 × 20

= 500m²

area of the triangle = 1/2bh

= 1/2 × 10 × 25

= 25 × 5

= 125 m²

Therefore the area of the composite figure

= 125 + 500 + 175

= 800m²

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Answer:

a) All the students in Nadia's year are the population

b) All the students in Nadia's class are the sample

Step-by-step explanation:

She wants to find out if students in her year like the school lunches.

So, this is the population.

i.e All the students in Nadia's year

She asks her class to complete a survey,

So, this is the sample of the population (Her classmates are in the same year as her and so on)

i.e All the students in Nadia's class

Yes, we would have reason to believe that the proportion of rats developing tumors when subjected to this diet has increased if the experiment were repeated and 16 of 48 rats developed tumors.

To determine whether there is an increase in the proportion of rats developing tumors when subjected to a coffee bean diet, we can conduct a hypothesis test using the 0.05 level of significance.

1. State the hypotheses:

  - Null hypothesis (H0): The proportion of rats developing tumors remains the same.

  - Alternative hypothesis (Ha): The proportion of rats developing tumors has increased.

2. Identify the test statistic:

  We will use a z-test to compare the observed proportion of rats developing tumors with the expected proportion.

3. Set the significance level:

  The significance level (α) is given as 0.05.

4. Collect data:

  In the original experiment, 25% of rats developed tumors. In the repeated experiment, 16 out of 48 rats developed tumors.

5. Compute the test statistic:

  The test statistic formula for comparing proportions is:

  z = (p - P) / sqrt(P(1-P)/n)

  where p is the observed proportion, P is the hypothesized proportion, and n is the sample size.

  Using the observed proportion (16/48 = 0.333), the hypothesized proportion (0.25), and the sample size (48), we can calculate the test statistic.

6. Determine the critical value:

  Since we are using a 0.05 level of significance and conducting a one-tailed test (Ha: >), we can find the critical value from the standard normal distribution table. The critical value for a 0.05 significance level is 1.645.

7. Make a decision:

  If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that the proportion of rats developing tumors has increased.

8. Calculate the test statistic:

  Plugging in the values into the formula, we calculate the test statistic:

  z = (0.333 - 0.25) / sqrt(0.25 * 0.75 / 48) = 1.404

9. Compare the test statistic and critical value:

  The test statistic (1.404) is less than the critical value (1.645).

10. Make a decision:

   Since the test statistic is not greater than the critical value, we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to conclude that the proportion of rats developing tumors has increased when subjected to this diet.

In summary, based on the given data and conducting a hypothesis test, we do not have reason to believe that the proportion of rats developing tumors has increased if the experiment were repeated and 16 of 48 rats developed tumors.

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