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Elanso [62]
2 years ago
7

The minute hand of a watch has a length of 1.4 cm. Find the; (a) distance moved by the tip of the minute hand from 8.13 am to 8.

49 am. (b) angle swept by the minute hand if the tip moves 8.8 mm (c)time it will bee when the minute hand sweeps the angle calculated in (b), if the minute hand started moving at 8.39 am
Mathematics
1 answer:
LenKa [72]2 years ago
8 0

Answer:

A). The distance covered= 5.2752 cm

B). Angle swept = 36°

C). The time = 8:45 am

Step-by-step explanation:

The watch is assumed to be a circle while the minute hand is assumed to be the radius

Radius = 1.4cm

A). From 8:13 to 8:49= 36 minutes

Recall, there is 60 minutes in a round watch.

The distance covered= 2πr*36/60

The distance covered= 2*3.14*1.4*0.6

The distance covered= 5.2752 cm

B). If distance covered is 8.8 mm

8.8 mm = 0.88cm

0.88= 2*3.14*1.4*(x/360)

0.88/(8.792)=x/360

0.1= x/360

0.1= x/360

36°= x

Angle swept = 36°

C). The time it will be if the watch started from 8:39 am and moved 36°

36/360= y/60

0.1= y/60

0.1*60= y

6 minutes= y

The time with be 8:39+6 minutes

The time = 8:45 am

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On her first day in a hospital, Kiri receives u1 milligrams (mg) of a therapeutic drug. The amount of the drug Kiri receives inc
Valentin [98]

Answer:

a. 21 = u1 + 6·d

b. 29 = u1 + 10·d

c. d = 2, u1 = 9

Step-by-step explanation:

a. The given parameters are;

The amount of therapeutic drug Kiri receives on her first at the hospital = u1 milligrams

The amount of drug increase received by Kiri each day = d

The amount of drug Kiri received on the seventh day = 21 mg

The amount of drug she received on the eleventh day = 29 mg

Therefore, we have an arithmetic progression with the formula for the nth term given as follows;

aₙ = a₁ + (n - 1)·d

Where;

a₁ = u1

n = The number of terms

Therefore, for the 7th day, the amount of drugs she receives, which is 21 milligrams, is given as follows;

a₇ = u1 + (7 - 1)·d = u1 + 6·d = 21

The equation for the amount of drugs she receives in terms of u1 and d on the seventh day is given as follows;

21 = u1 + 6·d

b. For the eleventh day, the amount of drugs she receives, which is 29 milligrams, is given as follows;

a₁₁ = u1 + (11 - 1)·d = u1 + 10·d = 29

Therefore, the equation for the amount of drugs she receives in terms of u1 and d on the  eleventh day is given as follows;

29 = u1 + 10·d

c. Therefore, we have two equations which are given as follows;

21 = u1 + 6·d................(1)

29 = u1 + 10·d..............(2)

Subtracting equation (1) from equation (2) gives;

29 - 21 = (u1 + 10·d) - (u1 + 6·d)

8 = 4·d

d = 8/4 = 2

d = 2

From equation (1), we have;

21 = u1 + 6·d = u1 + 6×2 = u1 + 12

21 = u1 + 12

21 - 12 = u1

∴ u1 = 9

d = 2, u1 = 9.

4 0
1 year ago
Given the trinomial 5x^2-2x-3, predict the type of solutions.
pishuonlain [190]

Consider the trinomial 5x^2-2x-3.

1. Find the discriminant:

D=b^2-4ac=(-2)^2-4\cdot 5\cdot (-3)=4+60=64,\\ \\\sqrt{D}=8>0.

2. Find the roots of given trinomial:

x_{1,2}=\dfrac{-b\pm \sqrt{D}}{2a}=\dfrac{2\pm 8}{2\cdot 5}=1,-0.6.

3. Conclusion: given trinomial has two different real roots.

3 0
1 year ago
Read 2 more answers
We're testing the hypothesis that the average boy walks at 18 months of age (H0: p = 18). We assume that the ages at which boys
marusya05 [52]

Answer:

II. This finding is significant for a two-tailed test at .01.

III. This finding is significant for a one-tailed test at .01.

d. II and III only

Step-by-step explanation:

1) Data given and notation    

\bar X=19.2 represent the battery life sample mean    

\sigma=2.5 represent the population standard deviation    

n=25 sample size    

\mu_o =18 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)    

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean battery life is equal to 18 or not for parta I and II:    

Null hypothesis:\mu = 18    

Alternative hypothesis:\mu \neq 18    

And for part III we have a one tailed test with the following hypothesis:

Null hypothesis:\mu \leq 18    

Alternative hypothesis:\mu > 18  

Since we know the population deviation, 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".    

3) Calculate the statistic    

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

z=\frac{19.2-18}{\frac{2.5}{\sqrt{25}}}=2.4    

4) P-value    

First we need to calculate the degrees of freedom given by:  

df=n-1=25-1=24  

Since is a two tailed test for parts I and II, the p value would be:    

p_v =2*P(t_{(24)}>2.4)=0.0245

And for part III since we have a one right tailed test the p value is:

p_v =P(t_{(24)}>2.4)=0.0122

5) Conclusion    

I. This finding is significant for a two-tailed test at .05.

Since the p_v. We reject the null hypothesis so we don't have a significant result. FALSE

II. This finding is significant for a two-tailed test at .01.

Since the p_v >\alpha. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

III. This finding is significant for a one-tailed test at .01.

Since the p_v >\alpha. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

So then the correct options is:

d. II and III only

6 0
2 years ago
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trapecia [35]

Answer:

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p + m = 19

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Step-by-step explanation:

To set up a system of equations, use the total number purchased as the first equation.

p + m = 19

Now use the cost and the total spent for the second equation.

0.25p + 0.75m = 11.50

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A)-2
9=8x+25
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X=-2
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