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ira [324]
3 years ago
6

What is the distance between point A and point B

Mathematics
1 answer:
Maru [420]3 years ago
5 0

Answer:

7x4

Step-by-step explanation:  180

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An Internet company sends and receives 582 e-mails every three seconds. How many e-mails does the company handle in one 24-hour
lara31 [8.8K]

We can set up a proportion 582/3=x/86400  (86400 is amount of seconds in a day)

Multiply 582 and 86400  You get 50284800

Divide that by 3 You get 16,761,600 emails in a 24 hour period

5 0
4 years ago
PICTURE BELOW / What is the volume of this cube?
Rama09 [41]

Answer:

74.088

or 74.09

Step-by-step explanation:

<h2>4.2³=</h2>

 4.2

 4.2

x4.2

74.088

<em>Rounded</em>

<h2>   ↓</h2><h3> 74.09</h3>
3 0
3 years ago
Read 2 more answers
I.5y=2x-5<br> II.5y=4+3x<br> III.5y-3x=-1
Ratling [72]

Answer:

Step-by-step explanation:

the vale of x differs in each situation are we talking AB value

6 0
3 years ago
The FDA regulates that a fish that is consumed is allowed to contain at most 1 mg/kg of mercury. In Florida, bass fish were coll
levacccp [35]

Answer:

Yes. At a significance level of 0.05, there is enough evidence to support the claim that the fish in all Florida lakes have different mercury than the allowable amount.

The random variable is the sample mean amount of mercury in the bass fish from the lakes of Florida.

The population parameter is the mean amount of mercury in the bass fish of Florida lakes.

The alternative hypothesis (Ha) states that the amount of mercury significantly differs from 1 mg/kg.

The null hypothesis (H0) states that the amount of mercury is not significantly different from 1 mg/kg.

H_0: \mu=1\\\\H_a:\mu\neq 1

Step-by-step explanation:

<em>The question is incomplete.</em>

<em>There is no data provided.</em>

<em>We will work with a sample mean of 0.95 mg/kg and sample standard deviation of 0.15 mg/kg to show the procedure.</em>

<em />

This is a hypothesis test for the population mean.

The claim is that the fish in all Florida lakes have different mercury than the allowable amount (1 mg of mercury per kg of fish).

Then, the null and alternative hypothesis are:

H_0: \mu=1\\\\H_a:\mu\neq 1

The significance level is assumed to be 0.05.

The sample has a size n=53.

The sample mean is M=0.95.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.15.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.15}{\sqrt{53}}=0.0206

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{0.95-1}{0.0206}=\dfrac{-0.05}{0.0206}=-2.427

The degrees of freedom for this sample size are:

df=n-1=53-1=52

This test is a two-tailed test, with 52 degrees of freedom and t=-2.427, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t

As the P-value (0.019) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the fish in all Florida lakes have different mercury than the allowable amount.

8 0
4 years ago
1. Sarah went to the farmers market because she needed 5 more apples for her Granny’s jam recipe. The recipe calls for 32 apples
ella [17]
5+x=32
x= the number of apples Sarah had before
5+x=32
subtract 5 from both sides
x=27
6 0
3 years ago
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