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3 years ago

If you are changing meters to centimeters, what direction do you move the decimal point?

1 answer:

3 0

to the right. think about it. there's 100 centimeters in a meter so you would want to move the decimal place in a way that increases the value you have.

ex: 5.76 meters. move to the right twice = 576 centimeters

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if Dakota earned $15.75 in interest in account A and $28.00 in interest in account B after 21 months. if the simple interest rat

NISA [10]

keeping in mind that 21 months is more than a year, since there are 12 months in a year, then 21 months is really 21/12 years.

$\bf ~~~~~~ \stackrel{\textit{account A}}{\textit{Simple Interest Earned}} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill &15.75\\ P=\textit{original amount deposited}\dotfill \\ r=rate\to 3\%\to \frac{3}{100}\dotfill &0.03\\ t=years\to \frac{21}{12}\dotfill &\frac{7}{4} \end{cases} \\\\\\ 15.75=P(0.03)\left( \frac{7}{4} \right)\implies \cfrac{15.75}{(0.03)\left( \frac{7}{4} \right)}=P\implies \boxed{300=P}$

$\bf ~~~~~~ \stackrel{\textit{account B}}{\textit{Simple Interest Earned}} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill &28\\ P=\textit{original amount deposited}\dotfill \\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ t=years\to \frac{21}{12}\dotfill &\frac{7}{4} \end{cases} \\\\\\ 28=P(0.049)\left( \frac{7}{4} \right)\implies \cfrac{28}{(0.049)\left( \frac{7}{4} \right)}=P\implies \boxed{326.53\approx P}$

so, clearly, you can see who's greater.

3 0

4 years ago

PLEASE HELP !!! ILL GIVE BRAINLIEST!! *DONT SKIP* ILL GIVE 40 POINTS.

Radda [10]

Answer:

it goes up by 9 and a half

4 0

3 years ago

Read 2 more answers

The equation y = 10x + 15 represents a gym that charges. What is the slope and y-intercept?

slega [8]

Answer:

Slope=10 Y=15

Step-by-step explanation:

Goes up by 10

Starts at 0,15

8 0

3 years ago

If the equation is (x + 1)(2x − 3) will the solution have two different lengths.

inysia [295]

Answer:

(x+1)(2x-3)

x(2x)=2x²

x(-3)=-3x

1(2x)=2x

1(-3)=-3

therefore

2x²-3x+2x-3

2x²-x-3

8 0

3 years ago

A student researcher compares the heights of American students and non-American students from the student body of a certain coll

OleMash [197]

Answer:

98% confidence interval for the true mean difference between the mean height of the American students and the mean height of the non-American students is [0.56 inches , 6.04 inches].

Step-by-step explanation:

We are given that a random sample of 12 American students had a mean height of 70.2 inches with a standard deviation of 2.73 inches.

A random sample of 18 non-American students had a mean height of 66.9 inches with a standard deviation of 3.13 inches.

Firstly, the Pivotal quantity for 98% confidence interval for the difference between the true means is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean height of American students = 70.2 inches

$\bar X_2$ = sample mean height of non-American students = 66.9 inches

$s_1$ = sample standard deviation of American students = 2.73 inches

$s_2$ = sample standard deviation of non-American students = 3.13 inches

$n_1$ = sample of American students = 12

$n_2$ = sample of non-American students = 18

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(12-1)\times 2.73^{2} +(18-1)\times 3.13^{2} }{12+18-2} }$ = 2.98

Here for constructing 98% confidence interval we have used Two-sample t test statistics.

So, 98% confidence interval for the difference between population means ( $\mu_1-\mu_2$ ) is ;

P(-2.467 < $t_2_8$ < 2.467) = 0.98 {As the critical value of t at 28 degree

of freedom are -2.467 & 2.467 with P = 1%}

P(-2.467 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.467) = 0.98

P( $-2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.98

P( $(\bar X_1-\bar X_2)-2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.98

98% confidence interval for ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

= [ $(70.2-66.9)-2.467 \times {s_p\sqrt{\frac{1}{12} +\frac{1}{18} } }$ , $(70.2-66.9)+2.467 \times {s_p\sqrt{\frac{1}{12} +\frac{1}{18} } }$ ]

= [0.56 , 6.04]

Therefore, 98% confidence interval for the true mean difference between the mean height of the American students and the mean height of the non-American students is [0.56 inches , 6.04 inches].

3 0

3 years ago