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2 years ago

Quently As

Mathematics

2 answers:

Pachacha [2.7K]2 years ago

8 0

We know that the sum of measure of any two sides of a triangle is greater than the third Angle.

So, in this case,

( a + b ) > c

Or, 9 + 10 > c

Or, 19 > c

Now, here only 19 is only greater than 18.3 from the options. Hence, 18.3 is the correct answer.

Anna71 [15]2 years ago

5 0

Answer:

A) 18.3

Step-by-step explanation:

Since adding b and a gives 19.

c must be less than 19.

So, A) 18.3 is the correct answer

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A pet store 145 bags of beef flavored dog food and 263 bag of chesse flavored dog food how many bags of dog food where sold

Kazeer [188]

Answer:

408 bags of dog food

Step-by-step explanation:

The question asks for the total bags of dog food. Because they give you two existing values expressing how much dog food of each type there is, all you must do is add them together. 145 + 263 = 408

3 0

3 years ago

Neil is a drummer who purchases his drumsticks online. When practicing with the newest pair, he notices they feel heavier than u

Fynjy0 [20]

Answer:

The probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.

Step-by-step explanation:

Given:

Weight of a given sample (x) = 2.33 oz

Mean weight (μ) = 1.75 oz

Standard deviation (σ) = 0.22 oz

The distribution is normal distribution.

So, first, we will find the z-score of the distribution using the formula:

$z=\frac{x-\mu}{\sigma}$

Plug in the values and solve for 'z'. This gives,

$z=\frac{2.33-1.75}{0.22}=2.64$

So, the z-score of the distribution is 2.64.

Now, we need the probability $P(x\geq 2.33 )=P(z\geq 2.64)$ .

From the normal distribution table for z-score equal to 2.64, the value of the probability is 0.9959. This is the area to the left of the curve or less than z-score value.

But, we need area more than the z-score value. So, the area is:

$P(z\geq 2.64)=1-0.9959=0.0041=0.41\%$

Therefore, the probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.

5 0

3 years ago

A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq

Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

$s=16.6$ represent the sample deviation

$\bar X=12.8$ represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

$\alpha=1-0.99=0.01$ represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

$\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.