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Talja [164]
2 years ago
8

A person drives 15km to the shops at an average speed of 30km/h, then drives back at an average speed of 60km/h. What is their a

verage speed over the two journeys?
Mathematics
1 answer:
____ [38]2 years ago
8 0
30 km/h to the store. 60 km/h back from the store.

(60 + 30) \ 2

We use the mean average formula.

90 \ 2

Final answer: 45
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You and a group of friends are going to a five-day outdoor music festival during spring break. You hope it does not rain during
KIM [24]

Answer:

The probability of not rain during the entire festival is 0.19

Step-by-step explanation:

The Complement Rule states that the sum of the probabilities of an event and its complement must equal 1.

In this case we have the probability of raining of each day,  we need the probability of NOT raining.  

First day= 45% chance of rain, the complement is 55%

Second day 55% chance of rain, the complement is 45%

Third day a 10% chance of rain, the complement is 90%

Fourth day a 10% chance of rain, the complement is 90%

Fifth day 5% chance of rain, the complement is 95%

To get the probability of NOT raining in the entire festival is the multiplication of all the complements.

P(not raining) = 0.55 x 0.45 x 0.90 x 0.90 x 0.95= 0.19

4 0
2 years ago
A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 449 gram setting. It is
stealth61 [152]

Answer:

We accetp  H₀

Step-by-step explanation:

Information:

Normal distribution  

Population mean      =   μ₀  = 449

Population standard deviation  σ   unknown

Sample size   n  =  23        n < 30    we use t-student test

so   n  =  23    degree of fredom   df = n  - 1  df  = 23- 1   df = 22

Sample mean    μ =  448

Sample standard deviation   s  =  20

Significance level  α  =  0,05  

1.-Hypothesis Test

Null hypothesis                               H₀     μ₀  =  449

Alternative hypothesis                    Hₐ     μ₀  ≠  449

Problem statement ask for determine decision rule for rejecting the null hypothesis. For rejecting the null hypothesis we have to  get an statistic parameter wich implies  that μ is bigger or smaller than μ₀

2.-Significance level   α  =  0,05  ;  as we have a two tail test

α/2    =  0,025

Then from t - student table for  df =  22   and 0,025 (two tail-test)

t(c)  =  ±  2.074

3.- Compute  t(s)

t(s)   =  (  μ  -  μ₀ )  /  s /√n

plugging in values

t(s)   =  (448  -  449) /  20 /√23    ⇒   t(s)   =  -  1*√23 /20

t(s)   =  - 0.2398

4.-Compare t(c)   and  t(s)

t(s)  <  t(c)         - 0.2398  <  - 2.074

Therefore  t(s)  in inside acceptance region.  We accept  H₀

7 0
2 years ago
Sam and Edna have 56 marbles together Edna has 6 times more marbles than Sam. How many marbles does sam have ?
Elis [28]
The answer is 8 marbles. 6*8=48. 48(ednas marbles) + 8(sam's marbles)=56.
7 0
2 years ago
Read 2 more answers
Jacob is solving the equation below using successive approximations.2^x-4=3^-x-2 He started from a graph where he found the solu
trasher [3.6K]

I think he made a mastake a step 3

4 0
2 years ago
Read 2 more answers
Describe the graph represented by the equation r=5/(3+2 sin theta).
Yuki888 [10]
We have the following equation:

r= \frac{5}{3+2sin(\theta)}

If we graph this equation we realize that in fact this is an ellipse with major axis matching the y-axis. So we can recognize these characteristics:

1. Center of the ellipse: 

The midpoint C<span> of the line segment joining the foci is called the </span>center<span> of the ellipse. So in this exercise this point is as follows:
</span>
C(0, -2)

2. Length of major axis:

The line through the foci is called the major axis<span>, so in the figure if you go from -5, at the y-coordinate, and walk through this major axis to the coordinate 1, the distance you run is the length of the major axis, that is:</span>

6 \ units

3. Length of minor axis:

The line perpendicular to the foci through the center is called the minor axis. So in the figure if you go from -2, at the x-coordinate, and walk through this minor axis to the coordinate 2, the distance you run is the length of the minor axis, that is:

4 \ units

4. Foci:

Let's find c as follows:

c=\sqrt{a^{2}-b^{2}}=\sqrt{3^{2}-2^{2}}=\sqrt{5}

Then the foci are:

f_{1}=(0, \sqrt{5}-2)

f_{2}=(0, -\sqrt{5}-2)

8 0
2 years ago
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