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beks73 [17]
2 years ago
8

Can someone help me for this?​

Mathematics
1 answer:
Ludmilka [50]2 years ago
8 0

Answer:

500000N=2m

x. =6m

cross multiply, it'll be

6×50000÷2

=150,000N

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What is the center of the data?
satela [25.4K]

Answer:

Median = 13

Step-by-step explanation:

Given

See attachment for dot plot

Required

The center of data

This implies that, we calculate the median

From the attached plot,

n = 13 --- number of observation

So, the median is:

Median = \frac{n + 1}{2}

Median = \frac{13 + 1}{2}

Median = \frac{14}{2}

Median = 7th

The median is at the 7th position

At the  7th position, is 13

Hence:

Median = 13

5 0
3 years ago
Read 2 more answers
13 - 3/2<br> Pls give an explanation
postnew [5]

13-3/2 you need to make a common denominator, so

26/2-3/2 then subtract the numerators, the denominator remains.

23/2 now simplify the fraction

11½ this is because 22/2+1/2=23/2, and 22/2=11, so 11+1/2 or 11½

7 0
3 years ago
Help me with my homework pls :)
vesna_86 [32]

Answer:

That smile tho lol

anywayssssss

b= 16 mm

Step-by-step explanation:

Pythagorean theorem is a^2 + b^2= c^2, so-

12^2 +b^2= 20^2

144+b^2=400

b^2=400-144

b^2= 256

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8 0
3 years ago
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What is the GCF of <br><br> 2X^3+3X^2+8X+12
tia_tia [17]

Answer:

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Step-by-step explanation:

5 0
3 years ago
A randomized controlled study was designed to test whether regular drinking of cranberry juice can prevent the recurrence of uri
Kamila [148]

Answer:

1. The chi-squared statistic = 10.36

The degrees of freedom = 17

The p-value for the test = 0.89

2. The range of the p-value from the Chi squared table = 0.75 < p-value < 0.90

Step-by-step explanation:

1. The Chi squared test is given as follows;

\chi ^{2} = \sum \dfrac{\left (Observed - Expected  \right )^{2}}{Expected  }

Therefore,

                              UTI   No UTI    %     Total

Cranberry juice       8           42      84     50

Lactobacillus          19          30       61     49

Control                    18          30      60    50

The chi-squared statistic is given as follows;

\chi ^{2} = \dfrac{\left (8- 18\right )^{2}}{18} +  \dfrac{\left (42 - 30\right )^{2}}{30} = 10.36

The chi-squared statistic = 10.36

The degrees of freedom, df = 18 - 1 = 17 since the all of the expected count have a minimum value of 18

With the aid of the calculator we find the p value as p as follows;

p = 0.9 - \dfrac{10.36 - 10.085}{12.972 - 10.085} \times (0.9 - 0.75)

The p-value for the test = 0.89  

2. The range of the p-value from the Chi squared table is given as follows;

0.75 < p-value < 0.90.

5 0
3 years ago
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