Which one is the answer

I need help on Part B and C on question 2

kap26 [50]

Answer:

The answer to your question is

Part A. Archimedes grades 6 1/4 tests per day

Part B. 8 19/32 days

Part C. 6 26/29 days

Step-by-step explanation:

Part A

Total time = 6 2/5 days

Number of tests = 40 tests

Process

1.- Convert the mixed fraction to improper fraction

6 2/5 = (30 + 2) / 5 = 32/5

2.- Divide 40 by 32/5

40/1 / 32/5 = (40 x 5) / (32 x 1)

= 200 / 32

Simplify

100 / 16 = 50/8 = 25/4

3.- Convert 25/4 to mixed fractions

6

4 25

1

25/4 = 6 1/4

Archimedes grade 6 1/4 tests per day

Part B

15 more tests

Total time = 32/5

Total tests = 40 + 15 = 55

Process

1.- Divide 55 by 32/5

55 / 1 / 32 /5 = (55 x 5) / (32 x 1)

= 275 / 32

2.- Convert 275/32 to a mixed fraction

8

32 275

256

19

Result 8 19/32 days

Part C

1.- Divide 55 by 7.25

50 / 7.25 = 5000 / 725

6

725 5000

- 4350

650

Result 6 650/750 = 6 26/29

3 0

3 years ago

Please help how do I get this answer ?

miss Akunina [59]

Multiply $20 by 20% good luck

8 0

3 years ago

Read 2 more answers

50 points!!!!! Please!!!!! Sandra and Jenny spend a certain amount of money from their money box each month to buy plants. The t

nekit [7.7K]

Answer:

A) Function 2 shows a greater rate of change because Jenny spends $6 each month and Sandra spends $5 each month.

Step-by-step explanation:

The y-intercept for Sandra is 55 and the slope is -5. While Jenny's y-intercept is 50 and slope is -6.

6 is a greater number therefore causing Jenny's slope to decrease faster causing her to have less money at a much faster rate. (rate of change=slope) Jeny's slope is decreasing at a faster rate of change because she is spending more money everyday than Sandra is.

Hope this helps! Have a great day! :)

4 0

3 years ago

Express 6^1/3 in simplest radical form.

seropon [69]

<em>Greetings from Brasil</em>

From radiciation properties:

$\large{A^{\frac{P}{Q}}=\sqrt[Q]{A^P}}$

bringing to our problem

$\large{6^{\frac{1}{3}}=\sqrt[3]{6^1}}$

6 0

4 years ago

A theory predicts that the mean age of stars within a particular type of star cluster is 3.3 billion years, with a standard devi

Luden [163]

Answer:

Null hypothesis: $\mu \leq 3.3$

Alternative hypothesis: $\mu > 3.3$

$z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768$

Since is a one right tailed test the p value would be:

$p_v =P(z>1.768)=0.039$

Step-by-step explanation:

Data given and notation

$\bar X=3.4$ represent the sample mean

$\sigma=0.4$ represent the population deviation for the sample

$n=50$ sample size

$\mu_o =3.3$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean is higher than 3.3, the system of hypothesis would be:

Null hypothesis: $\mu \leq 3.3$

Alternative hypothesis: $\mu > 3.3$

Compute the test statistic

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768$

P value

Since is a one right tailed test the p value would be:

$p_v =P(z>1.768)=0.039$

6 0

3 years ago