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jeka94
3 years ago
11

Complete the table of pork sandwiches to chicken

Mathematics
1 answer:
denis-greek [22]3 years ago
4 0

Step-by-step explanation:

Pork sandwiches: 2 , 6 , 10 , 14

goes by 4

Chicken Sandwiches: 5 , 15 , 25 , 35

goes by 10

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When sending a rectangular package through the U.S. Postal Service, the combined length and girth (perimeter of the cross sectio
weqwewe [10]
Let W = width of package
Let H = height of package
Let L = length of package

The perimeter cab be one of the following:
P = 2(L + W),  or
P = 2(L + H)

The perimeter of the cross section cannot exceed 108 in.
When the width is 10 in, then
 2(L + 10) <= 108
 L + 10 <= 54
 L <= 44 in

When the height is 15 in, then
2(L + 15) <= 108
L + 15 <= 54
L <= 39 in

To satisfy both of these conditions requires that L <= 39 in.

Answer: 39 inches
5 0
3 years ago
Given: cos α = 4/5, sin β= 4/5, and α and β are both in quadrant 1. find sin (α+β)
mezya [45]
We have the formula (sin x)^2 + (cos x)^2 = 1;
Then, sin α = \sqrt{1-  ( \frac{4}{5}) ^{2} } =  \frac{3}{5} ;
cos β = \sqrt{1- ( \frac{4}{5}) ^{2} } = \frac{3}{5} ;
We apply the formula sin ( α + β ) = sin α x cos β + sin β x cos α = (3/5)x(4/5) + (4/5)x(3/5) = 12/25 + 12/25 = 24/25;
8 0
3 years ago
What is the measure of 162<br> A. 324°<br> B. 162°<br> O C. 81°<br> D. 72°
telo118 [61]

Answer:

81

Step-by-step explanation:

trust me its correct

4 0
3 years ago
Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim
tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{10959}{25} =438.36

Sum of squares of differences = 175413.76

S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49

Population mean, μ = 425 mg/L

Sample mean, \bar{x} = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}

We use one-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813

Now, t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108

Since,                        

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0
3 years ago
(1 point) An innovative rural public health program is reducing infant mortality in a certain West African country. Pretend the
nataly862011 [7]

Answer:

Infant mortality will reduce to 33% in 1.63 years

Step-by-step explanation:

Let the population be X

Population after 1 year

X - 7.3 % of X

X -0.073 X\\0.927X

As we know

P = P_0 * e^{rt}

Substituting the given values we get -

\frac{P_0 - 0.33P_0}{P_0} = e^{-0.33 * t}\\0.67 = e^{-0.33 * t}\\

Taking log on both sides we get -

-0.20273 = -0.33 * t\\t = 1.627

4 0
3 years ago
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