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3 years ago

Ill give brainiest, dont guess please

Mathematics

2 answers:

IceJOKER [234]3 years ago

6 0

It could be private for the first one, if not I’m sorry

-The second national bank was chartered in 1816

andrew-mc [135]3 years ago

4 0

I don't know the first one but the second one is The second national bank was chartered in 1816.

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Please help ASAP !!!!!

sineoko [7]

Answer:

alright alright give us a second

27x to the 3rd power

3 0

3 years ago

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What is the simplified form of the ninth root of x times the ninth root of x times the ninth root of x times the ninth root of x

Ksivusya [100]

Since you are working with exponents and roots you would add the exponents together. The ninth root of x is equal to x^(1/9) and since you are using it four times your answer would be x^(4/9).

6 0

4 years ago

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Which value of x makes the numerical sentence true? 25^ 1/2 × 5 ^–3 = 5 ^x

Elenna [48]

5^-3=0.008
1/2=0.5
0.5*0.008=0.004
5/0.004=1250

5 0

4 years ago

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Let the vector \mathbf{v}v have an initial point at (1, -2)(1,−2) and a terminal point at (0, 0)(0,0). Plot the vector \mathbf{v

ASHA 777 [7]

Given that the initial point of the vector is (1,-2) and the termination point of the vector is (0,0).

So, the tail of the vector is at point (1,-2) and the head of the vector is at (0,0). The vector, $\vec{v}$ , has been shown in the figure.

The magnitude of the vector:

$|\vec{v}|=\sqrt{(1-0)^2+(-2-0)^2}=\sqrt{1+4}=\sqrt{5}$

The direction of a vector is the angle made by a vector with the positive direction of the x-axis.

From the figure, $\theta = 180 \degree\tan ^{-1} \left|\frac {0-(-2)}{0-1}\right|$

$\theta = 180 ^{\circ} - \tan^{-1}(2) = 180 ^{\circ} - 63.43^{\circ}$

$\theta =116.57 ^{\circ}$

Hence, the required vector having a magnitude $\sqrt {5}$ and direction $116.57 ^{\circ}$ has been shown in the figure.

4 0

4 years ago

The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili

Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 $s = \frac{\sigma}{\sqrt{n}}$ .