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3 years ago

Please help answer please

Mathematics

2 answers:

amid [387]3 years ago

8 0

Answer:

x = 19

Step-by-step explanation:

x-42+x-32+180-x+17 = 180

3x +123 = 180

3x = 57

x=19

WITCHER [35]3 years ago

3 0

x = 91.

Putting 91 into the two equations inside the triangle makes the angles 59 and 49, and makes the outside angle 108. Those two angles must add and equal the outside angle, and they do, meaning x = 91.

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The Lopez family and the Russell family each used their sprinklers last summer. The water output rate for the Lopez family's spr

grin007 [14]

L = hours used by the Lopez's sprinkler

R = hours used by the Russell's sprinkler

so, we know the Lopez's sprinkler uses 15 Liters per hour, so say after 1 hour it has used 15(1), after 2 hours it has used 15(2), after 3 hours it has used 15(3) liters and after L hours it has used then 15(L) or 15L.

likewise, the Russell's sprinkler, after R hours it has used 40R, since it uses 40 Liters per hour.

we know that both sprinklers combined went on and on for 45 hours, therefore whatever L and R are, L + R = 45.

we also know that the output on those 45 hours was 1050 Liters, therefore, we know that 15L + 40R = 1050.

$\bf \begin{cases}
L+R=45\implies \boxed{L}=45-R\\
15L+40R=1050\\
----------\\
15\left(\boxed{45-R} \right)+40R=1050
\end{cases}
\\\\\\
675-15R+40R=1050\implies 25R=375\implies R=\cfrac{375}{25}\\\\\\ R=15$

how long was the Lopez's on for? well, L = 45 - R.

7 0

4 years ago

A car is traveling at a rate of

LUCKY_DIMON [66]

50 meters per second

3600 seconds per hour

1000 meters per km

50/1000 = 0.05

1/3600 = .0002777

0.05/0.0002777 = 180km per hour

180*4 = 720km in 4 hours

8 0

4 years ago

Read 2 more answers

A restaurant chain has two locations in a medium-sized town and, believing that it has oversaturated the market for its food, is

scoray [572]

Answer:

Yes, since the test statistic value is greater than the critical value.

Step-by-step explanation:

Data given and notation

$\bar X_{1}=360000$ represent the mean for the downtown restaurant

$\bar X_{2}=340000$ represent the mean for the freeway restaurant

$s_{1}=50000$ represent the sample standard deviation for downtown

$s_{2}=40000$ represent the sample standard deviation for the freeway

$n_{1}=36$ sample size for the downtown restaurant

$n_{2}=36$ sample size for the freeway restaurant

t would represent the statistic (variable of interest)

$\alpha=0.05$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the true mean of revenue for downtown is higher than for freeway restaurant, the system of hypothesis would be:

Null hypothesis: $\mu_{1}\leq \mu_{2}$

Alternative hypothesis: $\mu_{1} > \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Determine the critical value.

Based on the significance level $\alpha=0.05$ and $\alpha/2=0.025$ we can find the critical values from the t distribution dith degrees of freedom df=36+36-2=55+88-2=70, we are looking for values that accumulates 0.025 of the area on the right tail on the t distribution.

For this case the value is $t_{1-\alpha/2}=1.66$

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{360000-340000}{\sqrt{\frac{50000^2}{36}+\frac{40000^2}{36}}}}=1.874$

What is the p-value for this hypothesis test?

Since is a right tailed test the p value would be:

$p_v =P(t_{70}>1.874)=0.033$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean for the downtown revenue restaurant seems higher than the true mean revenue for the freeway restaurant.

The best option would be:

Yes, since the test statistic value is greater than the critical value.

5 0

3 years ago

Find the coordinates of the midpoint M of the segment with the given endpoints. Then find the distance between the two points. R

Mrac [35]

Answer:

Midpoint (-2,4)

distance nearest tenth = 8.9

The approximate distance = 9

Step-by-step explanation:

Formulas

PQ midpoint = (x2 + x1)/2, (y2 + y1)/2

distance d = sqrt( (x2 - x1)^2 + (y2 - y1)^2 )

Givens

x2 = -4

x1 = 0

y2 = 1

y1 = 7

Solution

M(PQ) = (-4+0)/2, (1 + 7)/2

M(PQ) = -2, 4

The midpoint is -2,4

The distance = sqrt( (4 - 0)^2 + (1 + 7)^2 )

The distance = sqrt(16 + 64)

The distance = sqrt(80)

The distance = 4√5 exactly

The distance = 8.94

The distance = 8.9 To the nearest tenth

Question 2

The distance is rounded to the nearest whole number which is 9.

6 0

3 years ago

A heavy rope, 50 ft long, weighs 0.6 lb/ft and hangs over the edge of a building 120 ft high. Approximate the required work by a

Anastasy [175]

Answer:

Exercise (a)

The work done in pulling the rope to the top of the building is 750 lb·ft

Exercise (b)

The work done in pulling half the rope to the top of the building is 562.5 lb·ft

Step-by-step explanation:

Exercise (a)

The given parameters of the rope are;

The length of the rope = 50 ft.

The weight of the rope = 0.6 lb/ft.

The height of the building = 120 ft.

We have;

The work done in pulling a piece of the upper portion, ΔW₁ is given as follows;

ΔW₁ = 0.6Δx·x

The work done for the second half, ΔW₂, is given as follows;

ΔW₂ = 0.6Δx·x + 25×0.6 × 25 = 0.6Δx·x + 375

The total work done, W = W₁ + W₂ = 0.6Δx·x + 0.6Δx·x + 375

∴ We have;

W = $2 \times \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= 2 \times \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 750$

The work done in pulling the rope to the top of the building, W = 750 lb·ft

Exercise (b)

The work done in pulling half the rope is given by W₂ as follows;

$W_2 = \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 562.5$

The work done in pulling half the rope, W₂ = 562.5 lb·ft

6 0

3 years ago