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3 years ago

Find the value of x. Please help! Thanks.

Mathematics

1 answer:

emmasim [6.3K]3 years ago

8 0

Answer:

$\huge\boxed{x=\sqrt{30}}$

Step-by-step explanation:

ΔABC and ΔBDC are similar (AAA).

Therefore the corresponding sides are in proportion:

$\dfrac{AC}{BC}=\dfrac{BC}{CD}$

substitute

$AC=7+3=10\\{BC}=x\\CD=3$

$\dfrac{10}{x}=\dfrac{x}{3}$ cross multiply

$(10)(3)=(x)(x)\\\\30=x^2\\\\x^2=30\to x=\sqrt{30}$

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A large brine tank containing a solution of salt and water is being diluted with fresh water. The relationship between the elaps

kati45 [8]

Answer: 5.97 hours

Step-by-step explanation:

S(t) = 100

I put in 100 = 600 x e^-0.3t and it came out to be 5.97 when rounded to the nearest hundredth. It is also correct on khan academy.

7 0

3 years ago

Two major automobile manufacturers have produced compact cars with engines of the same size. We are interested in determining wh

Molodets [167]

Answer:

(A) The mean for the differences is 2.0.

(B) The test statistic is 1.617.

(C) At 90% confidence the null hypothesis should not be rejected.

Step-by-step explanation:

We are given that a random sample of eight cars from each manufacturer is selected, and eight drivers are selected to drive each automobile for a specified distance.

The following data (in miles per gallon) show the results of the test;

Driver Manufacturer A Manufacturer B

1 32 28

2 27 22

3 26 27

4 26 24

5 25 24

6 29 25

7 31 28

8 25 27

Let $\mu_1$ = mean MPG for the fuel efficiency of Manufacturer A brand

$\mu_2$ = mean MPG for the fuel efficiency of Manufacturer B brand

SO, Null Hypothesis, $H_0$ : $\mu_1-\mu_2=0$ or $\mu_1= \mu_2$ {means that there is a not any significant difference in the mean MPG (miles per gallon) when testing for the fuel efficiency of these two brands of automobiles}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2\neq 0$ or $\mu_1\neq \mu_2$ {means that there is a significant difference in the mean MPG (miles per gallon) for the fuel efficiency of these two brands of automobiles}

The test statistics that will be used here is 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 MPG for manufacturer A = $\frac{\sum X_A}{n_A}$ = 27.625

$\bar X_2$ = sample mean MPG for manufacturer B = $\frac{\sum X_B}{n_B}$ = 25.625

$s_1$ = sample standard deviation for manufacturer A = $\sqrt{\frac{\sum (X_A-\bar X_A)^{2} }{n_A-1} }$ = 2.72

$s_2$ = sample standard deviation manufacturer B = $\sqrt{\frac{\sum (X_B-\bar X_B)^{2} }{n_B-1} }$ = 2.20

$n_1$ = sample of cars selected from manufacturer A = 8

$n_2$ = sample of cars selected from manufacturer B = 8

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(8-1)\times 2.72^{2}+(8-1)\times 2.20^{2} }{8+8-2} }$ = 2.474

(A) The mean for the differences is = 27.625 - 25.625 = 2

(B) The test statistics = $\frac{(27.625-25.625)-(0)}{2.474 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }$ ~ $t_1_4$

= 1.617

(C) Now at 10% significance level, the t table gives critical values between -1.761 and 1.761 at 14 degree of freedom for two-tailed test. Since our test statistics lies within the range of critical values of t, 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 there is a not any significant difference in the mean MPG (miles per gallon) when testing for the fuel efficiency of these two brands of automobiles.

5 0

4 years ago

I have 10 dollars and my brother has 25 dollars. How much will I have if he gives me 15 dollars?

kotykmax [81]

you will have 25 dollars 10+15=25

4 0

4 years ago

Read 2 more answers

Plz hurry, brainliest and lots of points plz explain answer and only answer if you are actually gonna do it.

Citrus2011 [14]

Answer and Step-by-step explanation:

1st box:

0, -3

2nd box

-2, -9/4

3rd box:

2, -3/2

4th box:

4, 0

Btw 0, -2, 2, 4 are in the x-coordinate column if you didn’t know, and 1, -9/4, -3/2 are in the y-coordinate column

Basically to solve if you plug in 0 into x, so basically it’s 3(0)-4y=12, which you simplify gives you 0-4y=12, subtract 0 from 12 which is 12 and then do 12 divided by -4 and then you get y = -3 and that’s how you get the box answer

Idk how to the rest srry lol!

3 0

3 years ago

43x+ 67y = -24 and 67x + 43y = 24 solve

aleksandrvk [35]

Answer:

(x, y) = (1, -1)

Step-by-step explanation:

We'll write these equations in general form, then solve using the cross-multiplication method.

43x +67y +24 = 0

67x +43y -24 = 0

∆1 = (43)(43) -(67)(67) = -2640

∆2 = (67)(-24) -(43)(24) = -2640

∆3 = (24)(67) -(-24)(43) = 2640

These go into the relations ...

1/∆1 = x/∆2 = y/∆3

x = ∆2/∆1 = -2640/-2640 = 1

y = ∆3/∆1 = 2640/-2640 = -1

The solution is (x, y) = (1, -1).

_____

Additional comment

The cross multiplication method isn't taught everywhere. The attachment explains a bit about it. Our final relationship changes the order of the fractions to 1, x, y from x, y, 1. That way, we can use the equation coefficients in their original general-form order. (The fourth column in the 2×4 array of coefficients is a repeat of the first column.)

4 0

3 years ago