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1 year ago

Challenge A family wants to rent a car to go on vacation. Company A charges $70.50 and

Mathematics

1 answer:

lys-0071 [83]1 year ago

3 0

Answer:

7+×25

Step-by-step explanation:

I hope It help..........

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Which shows the graph of a line with the following slope and y-intercept?

poizon [28]

The answer is attachment 3 Because the y intercept is -5 and it goes down three and to the left 4.

8 0

2 years ago

Several programs attempt to address the shortage of qualified teachers by placing uncertified instructors in schools with acute

ss7ja [257]

Answer:

We conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

Step-by-step explanation:

We are given that reading scores of the students of certified teachers averaged 35.62 points with standard deviation 9.31. The scores of students instructed by uncertified teachers had mean 32.48 points with standard deviation 9.43 points on the same test.

There were 44 students in each group.

Let $\mu_1$ = mean scores with uncertified teachers.

$\mu_2$ = mean scores with certified teachers.

So, Null Hypothesis, $H_0$ : $\mu_1\geq \mu_2$ {means that the mean scores with uncertified teachers is higher or equal as compared to certified teachers}

Alternate Hypothesis, $H_A$ : $\mu_1$ {means that the mean scores with uncertified teachers is lower as compared to certified teachers}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviations;

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

where, $\bar X_1$ = sample mean scores of students instructed by uncertified teachers = 32.48 points

$\bar X_2$ = sample mean scores of students instructed by certified teachers = 35.62 points

$s_1$ = sample standard deviation of scores of students instructed by uncertified teachers = 9.43 points

$s_2$ = sample standard deviation of scores of students instructed by certified teachers = 9.31 points

$n_1$ = sample of students under uncertified teachers = 44

$n_2$ = sample of students under certified teachers = 44

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(44-1)\times 9.43^{2} +(44-1)\times 9.31^{2} }{44+44-2} }$ = 9.37

So, the test statistics = $\frac{(32.48-35.62)-(0)}{9.37 \times \sqrt{\frac{1}{44} +\frac{1}{44} } }$ ~ $t_8_6$

= -1.572

The value of t test statistics is -1.572.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical values of -1.665 at 86 degree of freedom for left-tailed test.

Since our test statistic is more than the critical values of t as -1.572 > -1.665, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

7 0

2 years ago

Randall has an account balance of $675.54 in a savings account that earns interest at a rate of 2.5% compounded twice a year. If

Alex_Xolod [135]

To calculate amount accrued after a given period of time we use the compound interest formula: A= P(1+r/100)∧n where A i the amount, P is the principal amount, r is the rate of interest and n is the interest period.
In the first part; A= $ 675.54, r= 1.25% (compounded semi-annually) and n =22 ( 11 years ), hence, 675.54 = P( 1.0125)∧22
= 675.54= 1.314P
P= $ 514.109 , therefore the principal amount was $ 514 (to nearest dollar)
Part 2
principal amount (p)= $ 541, rate (r) = 1.2 % (compounded twice a year thus rate for one half will be 2.4/2) and the interest period (n)= 34 (17 years×2)
Amount= 541 (1.012)∧34
= 541 ×1.5
= $ 811.5
Therefore, the account balance after $ 811.5.

5 0

2 years ago

Evaluate 2 tan 45 - tan 60

Neko [114]

Answer:

2-\sqrt{3} \\

Step-by-step explanation:

2-\sqrt{3} \\

4 0

2 years ago

Find the volume of a right circular cone that has a height of 11.8 ft and a base with a radius of 16 ft. Round your answer to th

andreyandreev [35.5K]

Answer:

3161.8 cubic foot

Step-by-step explanation:

V = 1/3BH = 1/3πr²H

V = 1/3 x 3.14 x 16² x 11.8 = 3161.77

4 0

2 years ago