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olasank [31]
4 years ago
5

Miss Lopez had 618 and 5400s in her checking account she deposits 50 and 2500 into her account how much will she have left in he

r account after she spends 47 and 2900s
Mathematics
1 answer:
34kurt4 years ago
8 0
618+5400=6018
6018+(50+2500)=8568
8568-(47+2900)=5621
she has $5,621 in her account
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Annette [7]
<h2>•5×7^2</h2>

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3 0
3 years ago
A coin contains 1.25 grams of nickel and 3.75 grams of copper, for a total weight of 5 grams. What percentage of the metal in th
otez555 [7]
So,

To find the percentage that the copper comprises, just divide the amount of copper by the total amount of metal in the coin.  In this case, we will change the denominator so that it is 100, which can easily be converted into a percentage, which is really just a fraction out of 100.

\frac{3.75}{5} = \frac{7.50}{10} = \frac{75}{100}

75%

75% of the metal in the coin is copper.
7 0
3 years ago
Which of the following is the solution to the system of equations?
drek231 [11]

Answer:

B (3/4,5/4)

Step-by-step explanation:

y = 2- x

\frac{5}{4}  = 2 -  \frac{3}{4}

y = 3x - 1

\frac{5}{4}  = 3 \times  \frac{3}{4}  - 1

\frac{5}{4}  =  \frac{9}{4}  -  \frac{4}{4}

Hope this helps ^-^

6 0
3 years ago
A dataset lists full IQ scores for a random sample of subjects with low lead levels in their blood (sample 1) and another random
Alenkasestr [34]

Answer:

a. Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 >\mu_2

b. t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282

c. p_v =P(t_{97}>2.287) =0.0122

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).  

Step-by-step explanation:

a. State and label the null and alternative hypotheses.

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 >\mu_2

Or equivalently:

Null hypothesis: \mu_1 - \mu_2 \leq 0

Alternative hypothesis: \mu_1 -\mu_2>0

Our notation on this case :

n_1 =78 represent the sample size for group 1

n_2 =21 represent the sample size for group 2

\bar X_1 =92.88 represent the sample mean for the group 1

\bar X_2 =86.90 represent the sample mean for the group 2

s_1=15.34 represent the sample standard deviation for group 1

s_2=8.99 represent the sample standard deviation for group 2

b. State the value of the test statistic.

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{s^2_2}{n_2}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom. If we replace the values given we have:

t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282

Now we can calculate the degrees of freedom given by:

df=78+21-2=97

c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )

Method used: P value

And now we can calculate the p value using the altenative hypothesis, since it's a right tail test the p value is given by:

p_v =P(t_{97}>2.287) =0.0122

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).  

6 0
4 years ago
If my dog is on a 5ft long leash, what is the area of the circle that he can run around in
xxMikexx [17]

Answer:

2/5

Step-by-step explanation:

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