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4 years ago

Solve each system of equations

Mathematics

1 answer:

vlada-n [284]4 years ago

5 0

Notice that we have -4y and 4y in the equations so we can just add them together to get:

6x-3x+4y-4y=3-2

3x=1

x=1/3, which makes -3x-4y=-2 become:

-1-4y=-2

-4y=-1

y=1/4

So the solution to the system of equations is the point:

(1/3, 1/4)

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Marissa created a graph to show her hair growth since her last haircut.

k0ka [10]

Answer:

A 2-column table with 3 rows titled Marissa's Hair Growth. Column 1 is labeled Months with entries 4, 6, 8. Column 2 is labeled Growth (inches) with entries 3, 4.5, 6.

Step-by-step explanation:

One of the given points on the graph is (4, 3), which corresponds to a table entry of Months = 4, Growth = 3.

Two tables have entries like that.

The points on the graph have a difference of 2 months, and a corresponding difference in growth of 1.5 inches. Only one of the two tables has entries like that. See below.

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3 years ago

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PLEASE ANSWER RIGHT AWAY

blsea [12.9K]

ANSWER

The second term is 7.

EXPLANATION

The given sequence has it's first term to be:

$t_1 = 3$

The recursive definition is :

$t_{n+1}=2t_n+n$

To find the second term , we substitute n=1,

to obtain,

$t_{1+1}=2t_1+1$

This implies that:

$t_{2}=2t_1+1$

$t_{2}=2(3)+1$

Simplify:

$t_{2}=6+1$

$t_{2} = 7$

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3 years ago

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One week in the summer Hanna worked 5 days. Each day she earned $2 more than the previous day. She earned a total of $60. How mu

grandymaker [24]

Answer:10$

Step-by-step explanation:

you have to times 5 and 2 u will get do$

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3 years ago

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Assume that foot lengths of women are normally distributed with a mean of 9.6 in and a standard deviation of 0.5 in.a. Find the

Makovka662 [10]

Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Step-by-step explanation:

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 $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be 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.

In this problem, we have that:

$\mu = 9.6, \sigma = 0.5$ .