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Ainat [17]
3 years ago
13

A function is given by the equation h (x)=x 2 squared +6x-3. what is h (3)?

Mathematics
1 answer:
Leno4ka [110]3 years ago
5 0
Your "<span>h (x)=x 2 squared +6x-3" is ambiguous.  If you meant "x squared," then you need only write x^2  OR  "x squared," but NOT "x 2 squared."

I will assume that by  "</span><span>h (x)=x 2 squared +6x-3" you actually meant:

</span><span>h(x)=x^2 +6x-3.  To find h(3), subst. 3 for x:  h(3) = (3)^2 + 6(3) - 3, or

h(3) = 9 + 18 - 3, or h(3) = 24.</span>
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1. A line goes through the points (4, 5) and (2, -6). Write the equation of the line in point-slope form. Show your work for ful
Hitman42 [59]
1. Find Slope
m = y2-y1/x2-x1
m = -6 - 5/2-4
m = -11/-2
m = 11/2

2. Write equation with know information
y = mx + b
y = 11/2x + b


3. Choose any of the two points (doesn’t matter which in you choose) and replace the x and y value into the equation
y = 11/2x + b (4,5)
5 = 11/2(4) + b
5 = 22 + b

Use algebra and solve for b
5 - 22 = b
b = -17

State all the information you know
m = 11/2 b = -17

Write the equation
y = 11/2x - 17
5 0
2 years ago
5. If John, travelling at a constant speed on a stretch of highway, covers 25 miles in 30
anzhelika [568]

Answer:

150 miles

Step-by-step explanation:

We can split 3 hours into six sections of 30 minutes, then multiply those six sections by the 25 miles. 6 times 25 equals 150

6 0
3 years ago
Find the range for the given data in 115, 534, 122, 599, 417, 289 A) 534 B) 167 C)115 D)484
Tamiku [17]
The answer is 484.Or D.
4 0
3 years ago
Read 2 more answers
Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and
TiliK225 [7]

Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33

What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

2.33 = \frac{X - 117}{4.33}

X - 117 = 2.33*4.33

X = 127.1

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

6 0
3 years ago
Last week, Jane made deposits of
liraira [26]

Answer:

b. $25

Step-by-step explanation:

64 + 25 + 37 - 52 - 49 = 25

6 0
3 years ago
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