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3 years ago

Alex defeated 10 gyms today! She started with 440 coins and now has 540 coins.

Mathematics

2 answers:

kap26 [50]3 years ago

6 0

Answer:

Are you asking how many coins she earned from each gym or-..? 10 gym do be crazy tho

SSSSS [86.1K]3 years ago

3 0

Answer:

good for Alex, she's got 100 more coins

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The area of a rhombus can be evaluated using the formula 1/2 d1d2 , where d1 represents the measure of one of its diagonals and

Eddi Din [679]

Answer:

Area of the rhombus will be a repeating decimal.

Step-by-step explanation:

In a terminating decimals, numbers get terminated after decimal like

1/4 = 0.25

while in repeating decimals, numbers get repeated after decimal like

1/3 = 0.33333

When we multiply two decimals which are repeating and terminating decimals the result will be a repeating decimal.

Therefore area of the rhombus will be a repeating decimal.

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2 years ago

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A high school has 2,000 students.

Rama09 [41]

Answer:

18 students

Step-by-step explanation:

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3 years ago

Which of the following can be a Pythagorean triplets?:

PolarNik [594]

<h2>Answer :-</h2>

As we know that,

Pythagoras triplet

1) a² + b² = c²

Let

$\sf \: {2}^{2} + {2}^{2} = {4}^{2}$

$\sf \: 4 + 4 = 16$

8 ≠ 16

<h3>Hence, A can't be Pythagoras triplet</h3>

2) a² + b² = c²

$\sf \:5 {}^{2} + {12}^{2} = {13}^{2}$

$\sf \: 25 + 144 = 169$

169 = 169

<h3>Therefore, B can be Pythagoras triplet</h3>

3)a² + b² = c²

$\sf {3}^{2} + {5}^{2} = {6}^{2}$

$\sf 9 + 25 = 36$

34 ≠ 36

<h3>Hence, C can't be Pythagoras triplet</h3>

4) a² + b² = c²

$\sf {5}^{2} + {4}^{2} = {7}^{2}$

$\sf \: 25 + 16 = 49$

41 ≠ 49

<h3>Hence, D can't be Pythagoras triplet</h3>

<h2 /><h2>Therefore :-</h2>

Only B can be Pythagoras triplet.

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2 years ago

The inequality 2x+2≤x+4≤3x+4 means that x is in the closed interval [A,B]

Alex_Xolod [135]

Answer:

Step-by-step explanation:

2x+2≤x+4≤3x+4

means : 2x+2≤x+4 ...(1) and x+4≤3x+4

(1) : 2x-x≤4-2 so : x≤2....(*)

(2) :3x+4 ≥x+4 .... 3x-x ≥ 0 so: x≥0 ....(**)

by (*) and (**) : 0≤x≤2 so : A=0 and B = 2

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2 years ago

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A study of 4,430 U.S. adults found that 2,913 were obese or overweight. The 95% confidence interval for the proportion of adults

IrinaK [193]

Answer:

$0.658 - 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.644$

$0.658 + 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.672$

We are 95% confident that the true proportion of adults overweight is between 0.644 and 0.672

Step-by-step explanation:

We can begin founding the estimated proportion of people overweight:

$\hat p =\frac{2913}{4430}= 0.658$

We need to find a critical value for the confidence level using the normal standard distribution. We know that 95% is the confidence level, then the significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the true population proportion of interest is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Replacing the data provided we got:

$0.658 - 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.644$

$0.658 + 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.672$

We are 95% confident that the true proportion of adults overweight is between 0.644 and 0.672

6 0

2 years ago