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Rudik [331]
2 years ago
9

Ben participated in a race. During the first half of the race, he walked 3 miles and ran at a rate of 5 miles per hour for x hou

rs. During the second half of the race, he walked for 2 miles and ran at a rate of 5.5 miles per hour for x hours. What equation could you write to solve for the number of hours Ben ran during each half of the race? Let y represent the number of miles in half the race. Write two equations, one for the first half of the race and one for the second half. How could you transform these two equations to be the same as the original equation you wrote to solve for x? What do you think this means about solving for the variables in a situation represented by two equations?
Mathematics
2 answers:
antiseptic1488 [7]2 years ago
8 0

Answer:

y = 5x + 3

y = 5.5x + 2

0.5x = 1

x = 2

y = 13

anyways what I think this means for the variables is x is equal to x in the second equation and y is equal to y in the second equation

Step-by-step explanation:

S_A_V [24]2 years ago
3 0
Y = 5x + 3
y = 5.5x + 2
0.5x = 1
x = 2
y = 13
I hope any of this will help you. c:
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When graphing the inequality y ≤ 2x − 4, the boundary line needs to be graphed first. Which graph correctly shows the boundary l
Aleksandr [31]

The answer is B.) 2nd. Picture

4 0
2 years ago
Kelly ordered 10 pizzas for a party. 30% of the pizzas have 12 slices each. The remaining 70% of the pizzas have 8 slices each.
abruzzese [7]

Answer:

The total number of slices is 92

Step-by-step explanation:

we know that

Kelly ordered 10 pizzas for a party

step 1

30% of the pizzas have 12 slices each

so

The number of pizzas for 30% is equal to

0.30(10)=3 pizzas

The number of slices is

3(12)=36 slices

step 2

70% of the pizzas have 8 slices each

so

The number of pizzas for 70% is equal to

0.70(10)=7 pizzas

The number of slices is

7(8)=56 slices

step 3

Find the total number of slices

36+56=92 slices

8 0
2 years ago
Which hyperbola has one focus point in common with the hyperbola (y+11)^2/(15^2)-(x-7)^2/(8^2)=1
irina [24]
\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)
\end{cases}\\\\
-------------------------------\\\\
\cfrac{(y+11)^2}{15^2}-\cfrac{(x-7)^2}{8^2}=1\implies \cfrac{(y-(-11))^2}{15^2}-\cfrac{(x-7)^2}{8^2}=1\\\\
-------------------------------\\\\
c=\textit{distance from the center to either foci}\\\\
c=\sqrt{a^2+b^2}\implies c=\sqrt{15^2+8^2}\implies \boxed{c=17}

now, if you notice, the positive fraction, is the one with the "y" variable, and that simply means, the hyperbola traverse axis is over the y-axis, so it more or less looks like the picture below.

now, from the hyperbola form, we can see the center is at (7, -11), and that the "c" distance is 17.

so, from -11 over the y-axis, we move Up and Down 17 units to get the foci, which will put them at (7, -11-17) or (7, -28) and (7 , -11+17) or (7, 6)

5 0
2 years ago
Read 2 more answers
7. The probability that a person living in a certain city owns a dog is estimated to be 0.3. Find the probability that the tenth
STALIN [3.7K]

Answer:lol can somebody please answer this I need an answer I have this question

Step-by-step explanation:

4 0
2 years ago
The blood platelet counts of a group of women have a​ bell-shaped distribution with a mean of 247.9 and a standard deviation of
labwork [276]

Answer:

A) Approximate percentage of women with platelet counts within 2 standard deviations of the​ mean, or between 118.5 and 377.3 = 95%

B) approximate percentage of women with platelet counts between 53.8 and 442.0 = 99.7%

Step-by-step explanation:

We are given;

mean;μ = 247.9

standard deviation;σ = 64.7

A) We want to find the approximate percentage of women with platelet counts within 2 standard deviations of the​ mean, or between 118.5 and 377.3.

Now, from the image attached, we can see that from the empirical curve, the probability of 1 standard deviation from the mean is (34% + 34%) = 68 %.

While probability of 2 standard deviations from the mean is (13.5% + 34% + 34% + 13.5%) = 95%

Thus, approximate percentage of women with platelet counts within 2 standard deviations of the​ mean, or between 118.5 and 377.3 = 95%

B) Now, we want to find the approximate percentage of women with platelet counts between 53.8 and 442.0.

53.8 and 442.0 represents 3 standard deviations from the mean.

Let's confirm that.

Since mean;μ = 247.9

standard deviation;σ = 64.7 ;

μ = 247.9

σ = 64.7

μ + 3σ = 247.9 + 3(64.7) = 442

Also;

μ - 3σ = 247.9 - 3(64.7) = 53.8

Again from the empirical curve attached, we cans that at 3 standard deviations from the mean, we have a percentage probability of;

(2.35% + 13.5% + 34% + 34% + 13.5% + 2.35%) = 99.7%

5 0
2 years ago
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