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2 years ago

What similarity statement can you write relating the three triangles in the diagram?

Mathematics

1 answer:

GarryVolchara [31]2 years ago

8 0

Answer:

They are both right-angled triangles.

Step-by-step explanation:

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Find the grade point average assume that the grade point values are 4.00 for an A,3.00 for a B,and so on B 2 B 4 B 3 C 3

VladimirAG [237]

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3 years ago

What is the BEST deal on sugar?

Lana71 [14]

Answer:

Step-by-step explanation:

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8 0

3 years ago

Write the equation of a line that passes through the point (-2, 1) and has a slope of 4.

stiv31 [10]

Answer:

y=4x-3

Step-by-step explanation:

y=(slope)x+b

y=4x+b

Insert Point:

$1=4(1)+b$

$b=-3$

3 0

3 years ago

The world's population is expected to grow at a rate of 1.3% per year until at least the year 2020. In 1994 the total population

Marina CMI [18]

The solution to your problem is as follows:

Given:
P₀ = 5,642×10^6 people,
i = 1.3% = 0.013 per year,
n = 2020 - 1994 = 26 years,

We get:

Pn = P₀ e^(i.n) 
. . .= 5,642×10^6×e^(0.013×26)
. . .= 7,911×10^6 people

Therefore, the the world's predicted population in the year 2020 is 7,911×10^6 people.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

5 0

3 years ago

Unlike most packaged food products, alcohol beverage container labels are not required to show calorie or nutrient content. An a

zhenek [66]

Answer:

We conclude that the true average estimated calorie content in the population sampled exceeds the actual content.

Step-by-step explanation:

We are given that an article reported on a pilot study in which each of 58 individuals in a sample was asked to estimate the calorie content of a 12 oz can of beer known to contain 153 calories.

The resulting sample mean estimated calorie level was 193 and the sample standard deviation was 88.

Let $\mu$ = true average estimated calorie content in the population sampled.

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 153 calories {means that the true average estimated calorie content in the population sampled does not exceeds the actual content}

Alternate Hypothesis, $H_A$ : $\mu$ > 153 calories {means that the true average estimated calorie content in the population sampled exceeds the actual content}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean estimated calorie level = 193 calories

s = sample standard deviation = 88

n = sample of individuals = 58

So, the test statistics = $\frac{193-153}{\frac{88}{\sqrt{58} } }$ ~ $t_5_7$

= 3.462

The value of t test statistics is 3.462.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 0.05 significance level the t table gives critical value of 1.6725 at 57 degree of freedom for right-tailed test.

Since our test statistic is more than the critical value of t as 3.462 > 1.6725, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the true average estimated calorie content in the population sampled exceeds the actual content.

8 0

3 years ago