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2 years ago

PLEASE HELP ME PLEASE WHICH EQUATION DOES THE GRAPH REPRESENT

Mathematics

1 answer:

noname [10]2 years ago

3 0

Answer:

Step-by-step explanation:

B is the best answer

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Is the relationship linear, exponential, or neither

photoshop1234 [79]

Answer: exponential

Step-by-step explanation:

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3 years ago

The mean for : $2.00$3.00$3.00$1.00$4.00$1.00$1.00$1.00 HELPPPPP PLSSSS

solmaris [256]

Answer:

Step-by-step explanation:

8 0

4 years ago

Order from least to greatest .9, 8/9, .86

labwork [276]

Answer:

least to greatest

.86, 8/9, .9

8 0

3 years ago

Read 2 more answers

One quart equals 946.35 milliliters. How many milliliters are in 3 pints

Allisa [31]

Answer:

There are 1419.5 ml in 3 pints

Step-by-step explanation:

The pint is a unit used to measure volume. When measuring many cups of liquid all put together we might want to use quarts.

A quart (qt) is the same thing as 4 cups or 2 pints.

1 quart = 2 pints = 4 cups

Now we have to find how many ml in 3 pints.

1 pint = 473.17647 ml

Now multiply the quantity of pint given in the question by 473.17647 ml.

3 * 473.17647

= 1419.52941 ml

By rounding off we get:

=1419.5 ml

Thus there are 1419.5 ml in 3 pints....

7 0

3 years ago

The following information was obtained from independent random samples taken of two populations. Assume normally distributed pop

Volgvan

Answer:

1. The 95% confidence interval for the difference between means is (-5.34, 11.34).

2. The standard error of (x-bar1)-(x-bar2) is 4.

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=10 has a mean of 45 and a standard deviation of √85=9.2195.

The sample 2, of size n2=12 has a mean of 42 and a standard deviation of √90=9.4868.

The difference between sample means is Md=3.

$M_d=M_1-M_2=45-42=3$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

The critical t-value for a 95% confidence interval is t=2.086.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=2.086 \cdot 4=8.34$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = 3-8.34=-5.34\\\\UL=M_d+t \cdot s_{M_d} = 3+8.34=11.34$

The 95% confidence interval for the difference between means is (-5.34, 11.34).

6 0

4 years ago