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Bogdan [553]
2 years ago
10

Parker is planning to build a playhouse for his sister.The scaled model below gives the reduced measures for width and height. H

ight 10 cm and base 22 cm. The yard space is large enough to have a playhouse that has a width of 3.5 meters. If Parker wants to keep the playhouse in proportion to the model, what cross multiplication of the proportion should he use to find the height?
Mathematics
2 answers:
kipiarov [429]2 years ago
6 0

Answer:

The cross multiplication would be 10(350) = 22(x).

Step-by-step explanation:

We would first convert meters to centimeters.  There are 100 cm in a meter; this means 3.5 m = 3.5(100) = 350 cm.

The ratio of the height to base of the model is 10/22.  The ratio of the playhouse would then be x/350.  This gives us

10/22 = x/350

Cross multiplying, we would have 10(350) = 22(x).

bezimeni [28]2 years ago
3 0

Answer:

its c (10)(3.5)=22x

Step-by-step explanation:

ed 2020

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2 years ago
Find the area of the triangle with the given measurements. Round the solution to the nearest hundredth if necessary. B = 105°, a
Akimi4 [234]

Answer:

The area of triangle is 110.10cm^{2}

Step-by-step explanation:

It is given that AB=19cm, BC=12cm and ∠B=105°

Now, Area of the triangle ABC=\frac{1}{2}acsinB

=\frac{1}{2}{\times}12{\times}19{\times}sin105^{\circ}

=6{\times}19{\times}(0.965)

=110.10cm^{2}

Thus, the area of triangle is 110.10cm^{2}

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Estimate the volume of the solid that lies below the surface z = xy and above the following rectangle. R = (x, y) | 2 ≤ x ≤ 8, 6
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2 years ago
A recent study reported that high school students spend an average of 94 minutes per day texting. Jenna claims that the average
larisa [96]

Answer:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

Step-by-step explanation:

Jenna claims that the average time of texting at her larger high school is greater than 94 minutes per day.

From here we can see that we have to perform a hypothesis test about a sample mean. The null and alternate hypothesis will be:

Null Hypothesis: \mu \leq  94

Alternate Hypothesis: \mu > 94

Jenna collected data from a sample of 32 students. So, sample size will be:

Sample Size = n = 32

Sample Mean = x = 96.5

Sample Standard Deviation = s = 6.3

We have to perform a hypothesis test, to test Jenna's claim. Since, the value of Population Standard Deviation is unknown and the value of Sample Standard Deviation is known, we will use One Sample t-test in this case.

The formula to calculate the test statistic is:

t=\frac{x-\mu}{\frac{s}{\sqrt{n}}}

Using the values, we get:

t=\frac{96.5-94}{\frac{6.3}{\sqrt{32} } }=2.245

The degrees of freedom will be:

df = n - 1 = 32 - 1 = 31

We have to convert the t-score 2.245 with 31 degrees of freedom to its equivalent p-value. From t-table this value comes out to be:

p-value = 0.0160

The significance level is:

\alpha =0.05

Since, the p-value is lesser than the level of significance, we reject the Null Hypothesis.

Conclusion:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

3 0
2 years ago
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