Answer:
7
Step-by-step explanation:
sing the equation > A = Pe^rt
20000 = 10000e^0.1t > 10% = 0.1
2 = e^0.1t
log 2 = log e^0.1t > log to the base e
log 2 = 0.1t * 1 log e to the base e =1
0.6931 = 0.1t
t = 6.931
= 7 years
Hope this helps <3
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<span>We use ratio and proportion to solve each of these:
</span><span>
</span><span>1.
The scale of a map is 1 in = 19.5 mi map: ________ in actual: 9.5 mi
</span><span>1 in / 19.5 mi = x in / 9.5 mi, x = 0.487 in
</span><span>
</span><span>2.
The scale of a map is 7 in = 16 mi map: 4.9 in actual: ______ mi
</span><span>7 in / 16 mi = 4.9 in / x mi, x = 11.2 mi
</span><span>
</span><span>3. The
scale factor for a model is 5 cm = ________ m Model : 72.5 cm actual:
165.3 m
</span><span>5 cm / x m = 72.5 cm / 165.3 m, x = 11.4 m
</span><span>
</span><span>4. The scale of a map is 1 in = 9.6 mi map: ________ in actual:
34.7 mi
</span><span>1 in / 9.6 mi = x in / 34.7 mi, x = 3.62 in
</span><span>
</span><span>5. The scale of a map is 1 ft = 9.6 mi map: ________ ft actual:
38.4 mi
</span><span>1 ft / 9.6 mi = x ft / 38.4 mi, x = 4 ft
</span><span>
</span><span>6. The scale factor for a model is 5 cm = ________ m Model :
22.4 cm actual: 155.2 m
</span><span>5 cm / x m = 22.4 cm / 155.2 m, x = 34.64 m
</span><span>
</span><span>7. The scale of a map is 5 in = 10 mi map: 8.7
in actual: ______ mi
</span><span>5 in / 10 mi = 8.7 in / x mi, x = 17.4 mi
</span><span>
</span><span>8. The scale of a map is 1 in = 13.5 mi map:
________ in actual: 65.9 mi
</span><span>1 in / 13.5 mi = x in / 65.9 mi, x = 4.88 in
</span><span>
</span><span>9. The scale factor for a model is 5 cm =
________ m Model : 61.5 cm actual: 143.2 m
</span><span>5 cm / x m = 61.5 / 143.2 m, x = 11.64 m
</span><span>
</span><span>10. The scale factor for a
model is 5 cm = ________ m Model : 29.7 cm actual: 179.5 m
</span><span>5 cm / x m = 29.7 cm / 179.5 m, x = 30.22 m
</span>
Answer:
The sum of the probabilities is greater than 100%; and the distribution is too uniform to be a normal distribution.
Step-by-step explanation:
The sum of the probabilities of a distribution should be 100%. When you add the probabilities of this distribution together, you have
22+24+21+26+28 = 46+21+26+28 = 67+26+28 = 93+28 = 121
This is more than 100%, which is a flaw with the results.
A normal distribution is a bell-shaped distribution. Graphing the probabilities for this distribution, we would have a bar up to 22; a bar to 24; a bar to 21; a bar to 26; and bar to 28.
The bars would not create a bell-shaped curve; thus this is not a normal distribution.
Example Problems<span>Find the value of "x": 5x = 25. Step 1 : The product is 25 and the given divisor is 5 . Given 5x = 25 . Divide by 5 on both the sides. 5 x 5 = 25 5 . ...<span>Find the value of "x" : 7x = 56; Step 1 : The product is 56 and the given divisor is 7. Given 7x = 56 . Divide by 7 on both the sides. 7 x 7 = 56 7 .</span></span>
Answer:
A) 34.13%
B) 15.87%
C) 95.44%
D) 97.72%
E) 49.87%
F) 0.13%
Step-by-step explanation:
To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:
Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:
So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:
P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)
= 0.5 - 0.1587 = 0.3413
It means that the PERCENT of scores that are between 90 and 100 is 34.13%
At the same way, we can calculated the percentages of B, C, D, E and F as:
B) Over 110
C) Between 80 and 120
D) less than 80
E) Between 70 and 100
F) More than 130
