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2 years ago

The number is greater than 40. The numberis NOT on a red mailbox. The number has more ones than tens.

Mathematics

1 answer:

UNO [17]2 years ago

4 0

Answer:

Step-by-step explanation:

Based on the clues given, Meg's mailbox number is equal to 89.

The clues adopted by Meg.

In this scenario, the three clues that were adopted by Meg include:

The number is greater than 40.

The number is NOT on a red mailbox.

The number has more ones than tens.

From the image, the numbers that are greater than 40 include:

Green Mail box = 60.

Red Mail Box = 58.

Blue Mail Box = 89.

Red Mail Box = 73.

From the image, the number that is not on a red mailbox are:

Green Mail box = 60.

Blue Mail Box = 89.

From the image, the number that has more ones than tens is:

Blue Mail Box = 89.

By using place value, 8 represents tens while 9 represents ones (unit). Therefore, Meg's mailbox number is equal to 89.

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Step-by-step explanation:

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3 years ago

Quickly solve (x+7)2=121, for x. Graph the solution on a number line.

asambeis [7]

Answer:

The value of the x is 53.5 or x=53.5

Step-by-step explanation:

Given data in the question:-

$(x+7)2=121$

We have to find the value of x.

Solution:-

$(x+7)2=121\\=>2x+14=121\\=>2x=121-14\\=>2x=107\\=>x=107/2\\=>x=53.5\\$

Hence the answer is 53.5

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3 years ago

Suppose that in a random selection of 100 colored candies, 21% of them are blue. The candy company claims that the percentage of

zhannawk [14.2K]

Answer:

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.

Step-by-step explanation:

The candy company claims that the percentage of blue candies is equal to 29%.

This means that the null hypothesis is:

$H_{0}: p = 0.29$

We want to test the hypothesis that this is true, so the alternate hypothesis is:

$H_{a}: p \neq 0.29$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.29 is tested at the null hypothesis:

This means that $\mu = 0.29, \sigma = \sqrt{0.29*0.71}$

Suppose that in a random selection of 100 colored candies, 21% of them are blue.

This means that $n = 100, X = 0.21$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.21 - 0.29}{\frac{\sqrt{0.29*0.71}}{\sqrt{100}}}$

$z = -1.76$

Pvalue of the test:

The pvalue of the test is the probability of the sample proportion differing at least 0.21 - 0.29 = 0.08 from the population proportion, which is 2 multiplied by the pvalue of Z = -1.76.

Looking at the z-table, z = -1.76 has a pvalue of 0.0392

2*0.0392 = 0.0784

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.

5 0

3 years ago

Alberto is painting a large triangle on one wall of his bedroom. The base of the triangle is 4 ft, and the height is 8 ft. It ta

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4 0

3 years ago

2x^2 -7= 74 please show all work

skad [1K]

$2x^2-7=74\\
2x^2=81\\
x^2=\dfrac{81}{2}\\
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x=-\dfrac{9}{\sqrt2} \vee x=\dfrac{9}{\sqrt2}}\\
x=-\dfrac{9\sqrt2}{2} \vee x=\dfrac{9\sqrt2}{2}}\\$

8 0

3 years ago