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mezya [45]
3 years ago
12

What is the interest rate in A = 12,000(1 + 0.07/4)^(4)(4)

Mathematics
1 answer:
evablogger [386]3 years ago
7 0

a=(3•107^16)/(40•400^14)
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I’m having trouble finding angle DCE, I figured out the angles in the kite (I hope), how do I find the angle next to it, it does
Rasek [7]

Answer:

180 -angle BCD

Step-by-step explanation:

if you've found angle BCD you should be able to find angle DCE as angles on a straight line equal 180°

7 0
3 years ago
Figure ABCD is a square. Prove BD ≅ AC. Statements Reasons 1. ABCD is a square 1. given 2. ∠DAB, ∠ABC, ∠BCD, and ∠CDA are right
FromTheMoon [43]
A. all sides are congruent. :) 

4 0
3 years ago
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Translate into a variable expression.<br> fifteen more than one-half of the square of z
stellarik [79]

Answer:

15 +(1/2)X z²

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5 0
3 years ago
Which of the following inequalities matches the graph?
Alik [6]
A vertical line has an equation of the form x = k, where k is the x-coordinate of all points on the line.
You have a vertical line. It passes through the point (-3, 0), so for this line, k = -3.
The vertical line has equation x = -3.
The line is dashed, not solid, so you have either < or >, but not <= or >=.
Also, notice the shading is to the left of x = -3, so all values of x are less than -3.
The inequality is

x < -3
8 0
3 years ago
A researcher wants to see if birds that build larger nests lay larger eggs. He selects two random samples of nests: one of small
True [87]

Answer:

95% confidence interval for the difference between the average mass of eggs in small and large nest is between a lower limit of 0.81 and an upper limit of 2.39.

Step-by-step explanation:

Confidence interval is given by mean +/- margin of error (E)

Eggs from small nest

Sample size (n1) = 60

Mean = 37.2

Sample variance = 24.7

Eggs from large nest

Sample size (n2) = 159

Mean = 35.6

Sample variance = 39

Pooled variance = [(60-1)24.7 + (159-1)39] ÷ (60+159-2) = 7619.3 ÷ 217 = 35.11

Standard deviation = sqrt(pooled variance) = sqrt(35.11) = 5.93

Difference in mean = 37.2 - 35.6 = 1.6

Degree of freedom = n1+n2 - 2 = 60+159-2 = 217

Confidence level = 95%

Critical value (t) corresponding to 217 degrees of freedom and 95% confidence level is 1.97132

E = t×sd/√(n1+n2) = 1.97132×5.93/√219 = 0.79

Lower limit = mean - E = 1.6 - 0.79 = 0.81

Upper limit = mean + E = 1.6 + 0.79 = 2.39

95% confidence interval for the difference in average mass is (0.81, 2.39)

3 0
3 years ago
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