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seraphim [82]
3 years ago
6

Based on a smartphone​ survey, assume that ​58% of adults with smartphones use them in theaters. In a separate survey of 225 adu

lts with​ smartphones, it is found that 114 use them in theaters.
a. If the 58​% rate is​ correct, find the probability of getting 114 or fewer smartphone owners who use them in theaters.
b. Is the result of 114 significantly​ low?
Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
8 0

Answer:

b

Step-by-step explanation:

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A cardboard sheet is cut in the shape of a triangle, with vertices at (0,0), (20,0), and (3,4) units. the thickness of the sheet
kenny6666 [7]
This problem can be solved in two ways, the long way, or the short way.

1. The long way
We know that the base of the triangle is along the x-axis, and the length of the base is 20.
The centre of mass is located at 2/3 of the distance from vertex (3,4) along the median, which cuts the base at (10,0).
Therefore the centre of mass is located at
x=3+(10-3)*2/3=23/3
y=4/3

2. The short way
It turns out that the centre of mass of a triangle sheet is located at the mean of the coordinates of the three vertices, i.e.
CG=((0+20+3)/3, (0+0+4)/3)=(23/3, 4/3)  as before.
6 0
3 years ago
Name a fraction equal to 1/2 with a denominator of 9
kolezko [41]
4.5/9. It is technically an improper fraction but it still counts.
3 0
3 years ago
Read 2 more answers
Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
tigry1 [53]

Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0
3 years ago
2 decreased by the sum of m and 8
Elina [12.6K]

Answer:

(m+8) - 2

Step-by-step explanation:

6 0
2 years ago
Write an expression to find the perimeter of<br> the rectangle below:<br> 2x - 3y<br> 3x - 1
LuckyWell [14K]

Answer:

2(5x-3y-1)

Step-by-step explanation:

perimeter of a rectangle formula:

2l+2w

l=length(2x-3y)5x

w=width(3x-1)

2(2x-3y)+2(3x-1)

4x-6y+6x-2

10x-6y-2

2) all of these numbers have a gcf of 2 so lets divide this equation by 2:

2(5x-3y-1)

hope this helps!

4 0
4 years ago
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