Estimate the product or quotient 4/7×1/6

Some of the animals that you can ride on the jungle carousel are lions,

Advocard [28]

4 cuz 6-2=4 sjsu i’d

5 0

3 years ago

A random sample of 17 hotels in Boston had an average nightly room rate of $165.40 with a sample standard deviation of $21.70. T

Citrus2011 [14]

Answer:

•Lower confidence limit = $155.2109

•Upper confidence limit ,= $174.5891

Step-by-step explanation:

We are given:

U = 17

x' = $165.40

S.d = $21.70

a = 10℅ (90℅ confidence interval)

For s.d(x'), we have:

S/√u = $\frac{21}{\sqrt{7}}$

= 5.2630

Therefore, S.E (x') = $5.2630

At a= 0.10 (90℅ confidence level)

and degree of freedom = 16 (17-1)

$Thus, t_1_6_, _0_._1_0 = 1.746$

90℅ confidence interval all around sample mean, will be:

(x' ± $t_(_n_-_1)$ ; a S.E (x')

= (165.4±(174.6)(5.2630)

(165.5±9.189)

($156.2109, $175.5891)

5 0

3 years ago

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2х5y-3z-=14 x-2y+=-12 -x+ Зу-2z =13

lukranit [14]

Answer:

x = -34/3, y = 1/3, and z = -35/3

Step-by-step explanation:

2х+5y-3z-=14 let's call this equation 1.

x-2y=-12 let's call this equation 2.

-x+Зу-2z =13 let's call this equation 3.

USING EQUATION 1 AND EQUATION 3.

2х+5y-3z-=14 (EQUATION 1)

-x+Зу-2z =13 (EQUATION 3)

LET'S GET RID OF z, BY MAKING THE COEFFICIENT OF z IN THE TWO EQUATION THE SAME. WE WILL MULTIPLY EQUATION 1 BY 2 AND EQUATION 3 BY 3.

2х+5y-3z-=14 (EQUATION 1) * 2

-x+Зу-2z =13 (EQUATION 3) * 3

4х+10y-6z-=28 let's call this equation 4.

-3x+9у-6z =39 let's call this equation 5.

TO GET RID OF z, WE WILL HAVE TO SUBTRACT EQUATION 5 FROM EQUATION 4 PLEASE TAKE NOTE OF THE SIGNS (-) (+).

4х+10y-6z-=28 (EQUATION 4)

- (-3x+9у-6z =39) (EQUATION 5)

(4x - 3x) + ((+10y) - (+9y)) + ((-6z) - (-6z)) = (28 - 39)

x + y + 0 = -11

x + y = -11 let's call this equation 6.

USING EQUATION 2 AND EQUATION 6, LET'S FIND x AND y.

x-2y=-12 (EQUATION 2)

x + y = -11 (EQUATION 6)

x has the same coefficient in both equations, which is 1. Let's get rid of x so we can find the value of y.

We will subtract equation 6 from equation 2. Take note of the signs.

x-2y=-12 (EQUATION 2)

- (x + y = -11) (EQUATION 6)

(x - x) + ((-2y) - (+y)) = ((-12) - (-11))

0 + -3y = -1

-3y = -1

Divide both sides by -3.

-3y/-3 = -1/-3

y = 1/3

LET'S SUBSTITUTE THE VALUE OF y INTO EQUATION 2 TO GET THE VALUE OF x.

x-2y=-12 (EQUATION 2)

x-2(1/3)=-12

x -2/3 = -12

Add 2/3 to both sides.

x -2/3 + (2/3) = -12 + (2/3)

x + 0 = -34/3

x = -34/3

LET'S SUBSTITUTE THE VALUES OF x and y INTO EQUATION 1 TO GET THE VALUE OF z.

2х+5y-3z-=14 (EQUATION 1)

2(-34/3)+5(1/3)-3z-=14

-68/3 + 5/3 - 3z = 14

-21 - 3z = 14

Add 21 to both sides.

-21 + (21) - 3z = 14 + (21)

0 - 3z = 35

-3z = 35

Divide both sides by -3.

-3z/-3 = 35/-3

z = -35/3 or

Therefore, x = -34/3, y = 1/3, and z = -35/3

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6 0

3 years ago

Triangle DEF (not shown) is similar to ABC shown, with angle B congruent to angle E and angle C congruent to angle F. The length

Bezzdna [24]

Answer:

Area of ΔDEF is $45\ cm^2$ .

Step-by-step explanation:

Given;

ΔABC and ΔDEF is similar and ∠B ≅ ∠E.

Length of AB = $2\ cm$ and

Length of DE = $6\ cm$

Area of ΔABC = $5\ cm^2$

Solution,

Since, ΔABC and ΔDEF is similar and ∠B ≅ ∠E.

Therefore,

$\frac{Area\ of\ triangle\ 1}{Area\ of\ triangle\ 2} =\frac{AB^2}{DE^2}$

Where triangle 1 and triangle 2 is ΔABC and ΔDEF respectively.

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

$\frac{5}{Area\ of\ triangle\ 2} =\frac{2^2}{6^2}\\ \frac{5}{Area\ of\ triangle\ 2}=\frac{4}{36}\\ Area\ of\ triangle\ 2=\frac{5\times36}{4} =5\times9=45\ cm^2$

Thus the area of ΔDEF is $45\ cm^2$ .

4 0

3 years ago

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Find the value of each variable. PHOTO ABOVE ⚠️TEST⚠️

telo118 [61]

Answer:

34

Step-by-step explanation:

5 0

3 years ago

Estimate the product or quotient4/7×1/6​

Estimate the product or quotient4/7×1/6