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2 years ago

Are the two triangles congruent ? Please answer correctly !!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!

Mathematics

2 answers:

mel-nik [20]2 years ago

4 0

Answer:

yes

Step-by-step explanation:

MrMuchimi2 years ago

3 0

Answer:

Step-by-step explanation:

None of the given sides or angles are congruent, so they are not congruent.

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Complete the pattern, 1.1, 2.2, 3-3. What is the rule?

spin [16.1K]

You mean 3.3? If that is the case then:

an=1.1+1(n-2)

or an=an-1+1
a1= 1.1

5 0

3 years ago

Vince added 0.912,0.882 ,and 1.354 liters of liquid to a bottle.About how much is in the bottle?

Levart [38]

3.148 liters of liquid.

3 0

3 years ago

When obtaining a confidence interval for a population mean in the case of a finite population of size N and a sample size n whic

Elanso [62]

Answer:

95% Confidence interval for the mean

$142.8 \leq\mu\leq158.4$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the mean of a finite population.

The error is multiplied by the following finite population correction factor:

$cf=\sqrt{\frac{N-n}{N-1} }$

The standard deviation can be estimated as

$\sigma=\frac{s}{\sqrt{n}} \sqrt{\frac{N-n}{N-1} } =\frac{24.4}{\sqrt{32} }* \sqrt{\frac{200-32}{200-1} }=3.963$

The 95% confidence interval has a z value of 1.96, so it becomes:

$M-z*\sigma_c\leq\mu\leq M+z*\sigma_c\\\\150.6-1.96*3.963\leq\mu\leq 150.6+1.96*3.963\\\\ 142.8 \leq\mu\leq 158.4$

5 0

3 years ago

In a positive correlation, __________.

Brut [27]

Im pretty sure its b

7 0

2 years ago

A Simple random sample of 100 8th graders at a large suburban middle school indicated that 84% of them are involved with some ty

Setler [38]

Answer: a) (0.755, 0.925)

Step-by-step explanation:

Let p be the population proportion of 8th graders are involved with some type of after school activity.

As per given , we have

n= 100

sample proportion: $\hat{p}=0.84$

Significance level : $\alpha= 1-0.98=0.02$

Critical z-value : $z_{\alpha/2}=2.33$ (using z-value table)

Then, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-

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

i.e. $0.84\pm (2.33)\sqrt{\dfrac{0.84(1-0.84)}{100}}$

i.e. $\approx0.84\pm 0.085$

i.e. $(0.84- 0.085,\ 0.84+ 0.085)=(0.755,\ 0.925)$

Hence, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity : a) (0.755, 0.925)

3 0

3 years ago