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STALIN [3.7K]
3 years ago
5

Skip-count to find the unknown area. Write a multiplication sentence for each tiled rectangle. You can use any color below to co

mplete each tile.
Mathematics
1 answer:
Vikki [24]3 years ago
5 0

Answer:

(a) Length= 3cm   Width = 6cm   Area = 24cm^2

(b) Length= 3cm   Width = 8cm     Area= 24cm^2

Step-by-step explanation:

Given

See attachment for rectangle

Solving (a):

From the first rectangle, there are 6 columns and 3 rows.

This implies that:

Length= 3cm

Width = 6cm

So, the area is:

Area = Length * Width

Area = 3cm * 6cm

Area = 18cm^2

Solving (a):

From the first rectangle, there are 3 rows and the total area is 24.

This implies that:

Length= 3cm

Area= 24cm^2

Area is:

Area = Length * Width

24cm^2= 3cm * Width

Make Width the subject

Width = \frac{24cm^2}{3cm}

Width = 8cm

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The area of the kite is 60in^2. Find the value of x.
Mashcka [7]

Answer:

8 in

Step-by-step explanation:

Area = (d1 × d2)/2

d1 = 15 in

d2 = x in

60 = ( 15 * x)/2

120 = 15x

x = 120/15

= 8 in

6 0
3 years ago
The number 5/9 is an example of a
Bond [772]

Answer:

5/9 is an example of rational number.

Step-by-step explanation:

as the definition of rational number is the number in the form of p/q but the value of denominator q should not be equal to 0. Every integer is a rational number.

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3 years ago
Multiply the binomials:<br><br> (3x + 4)(5x − 2)
seraphim [82]

Answer:

15x^2 + 14x - 8

Step-by-step explanation:

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2 years ago
PLEASE HELP ASSSAAAPPPPP
Elis [28]
B. would be your answer, I'm almost 100% sure.

Brainliest?
It would mean a lot!
6 0
3 years ago
The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation
fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 197.5, \sigma = 8.3

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{210.9 - 197.5}{8.3}

Z = 1.61

Z = 1.61 has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{196 - 197.5}{2.14}

Z = -0.7

Z = -0.7 has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0
3 years ago
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