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Svetllana [295]

2 years ago

12

Lily build and orange shelf and a brown shelf to hang in her room the orange shelf is 5 feet 6inches long the brown shelf is twi

ce as long what is the length of the brown shelf

1 answer:

astra-53 [7]2 years ago

8 0

Answer: 10 feet and 12 inches

Step-by-step explanation:

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If a set of scores for a variable is displayed in a frequency distribution polygon, which scale of measurement was used to measu

liraira [26]

Given: The set of scores for a variable is displayed in a frequency distribution polygon.

To find: We need to find which scale of measurement was used to measure the variable

Solution: A frequency distribution table is always either in interval or in ration scale.

4 0

2 years ago

PLEASE HELP ME????( Brainlest) Please Show Work???

REY [17]

Step-by-step explanation:

again, 2 unknowns.

x = number of 2- points baskets

y = number of 3- points baskets

we know they hit the basket 37 times in a so far unknown mixture of 2-point and 3- point throws.

x + y = 37

which gives us e.g.

x = 37 - y

and we know that with this unknown mixture they scored 80 points.

so,

2x + 3y = 80

as every successful 2-points throw scores 2 points, and every successful 3-points throw scores 3 points.

so, again our 2 equations :

x = 37 - y

2x + 3y = 80

remember, we prepared the first equation so that it gives us already an identity expressing one variable by the other. and that we use in the second equation :

2×(37 - y) + 3y = 80

74 - 2y + 3y = 80

74 + y = 80

y = 6

and from

x = 37 - y

we get

x = 37 - 6 = 31

so, they had 31 2-points throws and 6 3-points throws.

3 0

1 year ago

June has a savings account with an annual simple interest rate of 2.6%. She hopes to gain $6,500 in interest over a period of el

leonid [27]

$\bf \qquad \textit{Simple Interest Earned}\\\\
I = Prt\qquad 
\begin{cases}
I=\textit{interest earned}\to &\$6500\\
P=\textit{original amount deposited}\\
r=rate\to 2.6\%\to \frac{2.6}{100}\to &0.026\\
t=years\to &3
\end{cases}$

she should have invested P, in order to get I = 6500

but she instead invested 17,801, how much more did she need? well, P - 17801

solve for P

7 0

3 years ago

Read 2 more answers

PLEASE HELP ILL GIVE 77 POINTS!!!!!!!!!!!!

Vikki [24]

Answer:

i think its twelve nine and five

Step-by-step explanation:

not one hundred percent sure

8 0

2 years ago

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a st

Alinara [238K]

Answer:

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

Step-by-step explanation:

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

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

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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so $\mu = 0.5, \sigma = 0.05$ .

What is the probability that a line width is greater than 0.62 micrometer?

That is $P(X > 0.62)$

So

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

$Z = \frac{0.62 - 0.5}{0.05}$

$Z = 2.4$

Z = 2.4 has a pvalue of 0.99180.

This means that P(X \leq 0.62) = 0.99180.

We also have that

$P(X \leq 0.62) + P(X > 0.62) = 1$

$P(X > 0.62) = 1 - 0.99180 = 0.0082$

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

3 0

2 years ago