College Algebra

Which answer shows 9x 10-5 written in standard form?

kolezko [41]

Answer:

0.0009

Step-by-step explanation:

8 0

3 years ago

How to find an exponential function with two points?

BabaBlast [244]

Well u have a basic equation y=a • b^x
a is the initial amount
b is the growth factor
and x is the exponent

the first step is to make a chart with ur two points for example
(0,4) and (2,16)
so ur chart will be
x. | y.
0. | 4
2. | 16
next you find the difference between the x side so 0 to 2 is +2
then find the difference between the y side so 4 to 16 is +12
then put it into a fraction with y over x or y/x so 12/2 then simplified 6/1 or just 6.

6 is the growth factor
and to find a u have to go on the y column and find the first number so
a is 4
x is still x because it's 0 but if it was 2 then it would be x-2 so it can cancel
so the answer would be y=4 • 6^x

6 0

2 years ago

Find the equation of a parabola with a vertical axis and its vertex at the origin and passing through the point (-2, 3)

vredina [299]

a vertical axis, I assume it means a vertical axis of symmetry, thus it'd be a vertical parabola, like the one in the picture below.

$\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \qquad \leftarrow vertical\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=0\\ k=0 \end{cases}\implies y=a(x-0)^2+0 \\\\\\ \textit{we also know that } \begin{cases} x=-2\\ y=3 \end{cases}\implies 3=a(-2-0)^2+0\implies 3=4a \\\\\\ \cfrac{3}{4}=a~\hspace{10em}y=\cfrac{3}{4}(x-0)^2+0\implies \boxed{y=\cfrac{3}{4}x^2}$

8 0

2 years ago

Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights

user100 [1]

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

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.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that $\mu = 85, \sigma = 8$

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

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

$Z = \frac{100 - 85}{8}$

$Z = 1.875$

$Z = 1.875$ has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

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

$-0.84 = \frac{X- 85}{8}$

$X - 85 = -0.84*8$

$X = 78.28$

The weight that 80% of the apples exceed is of 78.28g.

5 0

2 years ago

One stats class consists of 52 women and 28 men. Assume the average exam score on Exam 1 was 74 (σ = 10.43; assume the whole cla

Svetllana [295]

Answer:

(A) What is the z- score of the sample mean?

The z- score of the sample mean is 0.0959

(B) Is this sample significantly different from the population?

No; at 0.05 alpha level (95% confidence) and (n-1 =79) degrees of freedom, the sample mean is NOT significantly different from the population mean.

Step -by- step explanation:

(A) To find the z- score of the sample mean,

X = 75 which is the raw score

¶ = 74 which is the population mean

S. D. = 10.43 which is the population standard deviation of/from the mean

Z = [X-¶] ÷ S. D.

Z = [75-74] ÷ 10.43 = 0.0959

Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].

(B) To test for whether this sample is significantly different from the population, use the One Sample T- test. This parametric test compares the sample mean to the given population mean.

The estimated standard error of the mean is s/√n

S. E. = 16/√80 = 16/8.94 = 1.789

The Absolute (Calculated) t value is now: [75-74] ÷ 1.789 = 1 ÷ 1.789 = 0.559

Setting up the hypotheses,

Null hypothesis: Sample is not significantly different from population

Alternative hypothesis: Sample is significantly different from population

Having gotten T- cal, T- tab is found thus:

The Critical (Table) t value is found using

- a specific alpha or confidence level

- (n - 1) degrees of freedom; where n is the total number of observations or items in the population

- the standard t- distribution table

Alpha level = 0.05

1 - (0.05 ÷ 2) = 0.975

Checking the column of 0.975 on the t table and tracing it down to the row with 79 degrees of freedom;

The critical t value is 1.990

Since T- cal < T- tab (0.559 < 1.990), refute the alternative hypothesis and accept the null hypothesis.

Hence, with 95% confidence, it is derived that the sample is not significantly different from the population.

6 0

3 years ago