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nasty-shy [4]
2 years ago
7

Every weekday, Mr. Jones bikes from his home to his job. Sometimes he rides along two roads, the long route that is shown by the

solid lines. Other times, he takes the shortcut shown by the dashed line.
A right triangle. The distance from the angle with measure 90 degrees to Job is 15 kilometers. From the angle to Home is a. They hypotenuse is 17 kilometers.


How many fewer kilometers does Mr. Jones bike when he takes the shortcut instead of the long route?

6 kilometers

8 kilometers

23 kilometers

32 kilometers

Mark this and return

Mathematics
2 answers:
azamat2 years ago
8 0

Answer:

a

Step-by-step explanation:

on edge

notka56 [123]2 years ago
5 0

Answer:

(A)6 kilometers

Step-by-step explanation:

First, we determine the value of a using Pythagoras Theorem.

Hypotenuse^2=Opposite^2+Adjacent^2\\17^2=a^2+15^2\\a^2=17^2-15^2\\a^2=289-225\\a^2=64\\a^2=8^2\\a=8$ km

Therefore:

Distance along the long route = 8 + 15 =23 km

Distance along the shortcut =17 km

Difference =23-17 =6km

Therefore, Mr. Jones bikes 6km less when he takes the shortcut instead of the long route.

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Answer:

We can do it with envelopes with amounts $1,$2,$4,$8,$16,$32,$64,$128,$256 and $489

Step-by-step explanation:

  • Observe that, in binary system, 1023=1111111111. That is, with 10 digits we can express up to number 1023.

This give us the idea to put in each envelope an amount of money equal to the positional value of each digit in the representation of 1023. That is, we will put the bills in envelopes with amounts of money equal to $1,$2,$4,$8,$16,$32,$64,$128,$256 and $512.

However, a little modification must be done, since we do not have $1023, only $1,000. To solve this, the last envelope should have $489 instead of 512.

Observe that:

  1. 1+2+4+8+16+32+64+128+256+489=1000
  2. Since each one of the first 9 envelopes represents a position in a binary system, we can represent every natural number from zero up to 511.
  3. If we want to give an amount "x" which is greater than $511, we can use our $489 envelope. Then we would just need to combine the other 9 to obtain x-489 dollars. Since x-489\leq511, by 2) we know that this would be possible.

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3 years ago
6R + r -5r simplify each expression<br> A.2r<br> B.1 r + r<br> C.0r<br> D.7r-5r
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A survey of 700 adults from a certain region​ asked, "What do you buy from your mobile​ device?" The results indicated that 59​%
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Answer:

a) the test statistic z = 1.891

the null hypothesis accepted at 95% level of significance

b) the critical values of 95% level of significance is zα =1.96

c) 95% of confidence intervals are  (0.523 ,0.596)

Step-by-step explanation:

A survey of 700 adults from a certain region

Given sample sizes n_{1} = 400 and n_{2} = 300

Proportion of mean p_{1} = \frac{236}{400} = 0.59 and p_{2} = \frac{156}{300} = 0.52

<u>Null hypothesis H0</u> : assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

<u>Alternative hypothesis H1:</u>- p1 ≠ p2

a) The test statistic is

Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )}  } }

where p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}  

on calculation we get   p = 0.56    

now q =1-p = 1-0.56=0.44

Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )}  } }\\   =\frac{0.56-0.52}{\sqrt{0.56X0.44}(\frac{1}{400}+\frac{1}{300}   }

after calculation we get z = 1.891

b) The critical value at 95% confidence interval zα = 1.96 (from z-table)

The calculated z- value < the tabulated value

therefore the null hypothesis accepted

<u>conclusion</u>:-

assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

c) <u>95% confidence intervals</u>

The confidence intervals are P± 1.96(√PQ/n)

we know that = p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}

after calculation we get P = 0.56 and Q =1-P =0.44

Confidence intervals are ( P- 1.96(√PQ/n), P+ 1.96(√PQ/n))

now substitute values , we get

( 0.56- 1.96(√0.56X0.44/700), 0.56+ 1.96(0.56X0.44/700))

on simplification we get (0.523 ,0.596)

Therefore the population proportion (0.56) lies in between the 95% of <u>confidence intervals  (0.523 ,0.596)</u>

<u></u>

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3 years ago
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