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RoseWind [281]
4 years ago
11

Question 5!!!!!!!!!!!!!!!!

Mathematics
1 answer:
postnew [5]4 years ago
7 0

Answer:

median=59

Step-by-step explanation:

big brains for days

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JT bisects AJH and the measure of MJT is four times that of TJH. If the measures of MJT is 120 find the measure of MJA.
Ann [662]

MJH = 120

Let TJH = x

You’re told MJT is 4 times TJH

So you have 4x + x = 120

Simplify: 5x = 120

Divide both sides by 5:

X = 24

TJH = x = 24 degrees.

JT is a bisector so both TJH and TJA are the same. So TJA is also 24 degrees

So MJA = 120 - 24 - 24 = 72

MJA = 72 degrees

6 0
3 years ago
Read 2 more answers
5.1 is 102% of what number?
Cloud [144]

Answer:

5.202

Step-by-step explanation:

7 0
3 years ago
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(Based on Q1 ~ Q3) According to the Bureau of the Census, 18.1% of the U.S. population lives in the Northeast, 21.9% inn the Mid
vekshin1

Answer:

We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.

Step-by-step explanation:

The null Hypothesis: Geographical distribution of hotline callers could be the same as the U.S. population distribution

Alternative hypothesis: Geographical distribution of hotline callers could not be the same as the U.S. population distribution

The populations considered are the Midwest, South, Northeast, and west.

The number of categories, k = 4

Number of recent calls = 200

Let the number of estimated parameters that must be estimated, m = 0

The degree of freedom is given by the formula:

df = k - 1-m

df = 4 -1 - 0 = 3

Let the significance level be, α = 5% = 0.05

For  α = 0.05, and df = 3,

from the chi square distribution table, the critical value = 7.815

<u>Observed and expected frequencies of calls for each of the region:</u>

<u>Northeast</u>

Observed frequency = 39

It contains 18.1% of the US Population

The probability = 0.181

Expected frequency of call = 0.181 * 200 = 36.2

<u>Midwest</u>

Observed frequency = 55

It contains 21.9% of the US Population

The probability = 0.219

Expected frequency of call = 0.219 * 200 =43.8

<u>South</u>

Observed frequency = 60

It contains 36.7% of the US Population

The probability = 0.367

Expected frequency of call = 0.367 * 200 = 73.4

<u>West</u>

Observed frequency = 46

It contains 23.3% of the US Population

The probability = 0.233

Expected frequency of call = 0.233 * 200 = 46

x^{2} = \sum \frac{(O_{i} - E_{i})  ^{2} }{E_{i} } ,   i = 1, 2,.........k

Where O_{i} = observed frequency

E_{i} = Expected frequency

Calculate the test statistic value, x²

x^{2} = \frac{(39 - 36.2)^{2} }{36.2} + \frac{(55 - 43.8)^{2} }{43.8} + \frac{(60 - 73.4)^{2} }{73.4} + \frac{(46 - 46.6)^{2} }{46.6}

x^{2} = 5.535

Since the test statistic value, x²= 5.535 is less than the critical value = 7.815, the null hypothesis will not be rejected, i.e. it will be accepted. We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.  

7 0
3 years ago
Work out the gradient of a line that passes through co-ordinates (3,4) &amp; (6,16).<br>​
Ipatiy [6.2K]

Answer:

4.33

Step-by-step explanation:

Given data

x1= 3

y1=3

x2=6

y2=16

M= y2-y1/x2-x1

substitute

M=16-3/6-3

M= 13/3

M=4.33

Hence the gradient is 4.33

3 0
3 years ago
Find the difference quotient f(x+h)-f(x)/h , where, for the function below.
Assoli18 [71]
The difference quotient is -6. See photo for the work.

3 0
3 years ago
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