<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:
23
Step-by-step explanation:
we are given that angle NRQ is 78 degrees
we can see from the figure that the sum of the given angles is angle NRQ
so
(8x + 7) + (4x - 1) = 78
12x + 6 = 78
12x = 72
x = 6
Now, we have to find angle PRQ
replacing x with 6 in the equation of angle PRQ
PRQ = 4(6) - 1
PRQ = 23
Answer:
<em>The first step is to determine the average
</em>
<em>The exercise says it’s a normal distribution: (n=8)</em>
<em>According to the exercise, the mean is equal to 0,5 then the value of t of the distribution can be obtained
</em>
<em />
<em>The variable t has 7 grade to liberty, we calculate the p-value as:
</em>
This value is very high, therefore the hypothesis is not rejected
