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2 years ago

Use the net to find the size of the base and the height of each triangular face of this square pyramid.

Mathematics

2 answers:

nlexa [21]2 years ago

7 0

The base would be 16 and the height of each triangle would be 2

antiseptic1488 [7]2 years ago

4 0

The base is 16 and height of each triangle is 2

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Write and equation in slope intercept form of a line that has a slope of 1/3 and passes through the point (6,0)

ladessa [460]

y=(1/3)x-2

Step-by-step explanation:

the slope-intercept form is y-y1=m(x-x1)

hence, y-0=1/3(x-6) or y=(1/3)x-2

3 0

2 years ago

A veterinarian does research on the causes of enteroliths, stones that develop in the colon of horses. She suspects that feeding

dsp73

Answer:

On average, horses with and without enteroliths are fed the same amount of alfalfa per month

Step-by-step explanation:

Hello!

The veterinarian wants to research if eating alfalfa causes horses to have enteroliths.

The study variables are:

X₁: number of alfalfa flakes eaten over a month by a horse with enteroliths.

X₂: number of alfalfa flakes eaten over a month by a horse free of enteroliths.

She calculated a CI interval, with level 1 - α and the interval contained the cero.

With a significance level α, the hypotheses are:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

To decide whether you should reject or not the null hypothesis using a Confidence Interval is the following:

If the CI contains the number stated in the null hypothesis, the decision is to not reject the null hypothesis.

If the CI doesn't contain the number stated in the null hypothesis, then the decision is to reject the null hypothesis.

Since the interval contains the cero, if 1 - α and α are complementary, the decision is to reject the null hypothesis.

This means that at a level of significance of α, you can conclude that there is no difference between the population means of the numbers of alfalfa flakes eaten by horses with or without enteroliths.

I hope it helps!

7 0

2 years ago

What is the area of circle with a radius of 13 milimeters? use 3.14 for pi

MrRissso [65]

Answer:

530.66 mm²

5.3066 cm²

Step-by-step explanation:

A = π·r²

= 3.14×(13mm)²

= 3.14×169mm²

= 530.66 mm²

= 5.3066 cm²

8 0

2 years ago

Read 2 more answers

the barkery you choose costs $1,000 per month to rent. The bakery that you almost rented was 1/4 the cost but was much too small

liubo4ka [24]

Well the easiest way to figure out this question is to just divide it by 4, like quarters, so 1000 divided by 4 is 250. so the other bakery was only 250. hope this helped

4 0

2 years ago

Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The records of a medical clinic show

Dovator [93]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that there exist 60 or more defected children is $P(x \ge 60)=0.0901$

Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

in every 1000 children a particular genetic defect occurs to 1

The number of sample selected is $n= 50,000$

The probability of observing the defect is mathematically evaluated as

$p = \frac{1}{1000}$

$= 0.001$

The probability of not observing the defect is mathematically evaluated as

$q = 1-p$

$= 1-0.001$

$= 0.999$

The mean of this probability is mathematically represented as

$\mu = np$

Substituting values

$\mu = 50000*0.001$

$= 50$

The standard deviation of this probability is mathematically represented as

$\sigma = \sqrt{npq}$

Substituting values

$= \sqrt{50000 * 0.001 * 0.999}$

$= \sqrt{49.95}$

$= 7.07$

the probability of detecting $x \ge60$ defects can be represented in as normal distribution like

$P(x \ge 60)$

in standardizing the normal distribution the normal area used to approximate $P(x \ge 60)$ is the right of 59.5 instead of 60 because x= 60 is part of the observation

The z -score is obtained mathematically as

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

$= \frac{59.5 - 50 }{7.07}$

$=1.34$

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

The total area under the curve is 1

$P(x \ge 60) \approx P(z > 1.34)$

$= 1-P(z \le 1.34)$

$=1-0.9099$

$=0.0901$

3 0

2 years ago