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2 years ago

What is 3.405 written in expanded form

Mathematics

2 answers:

Eva8 [605]2 years ago

5 0

Not sure which expanded form you need it in. There's also expanded exponential form let me know if you need more assistance

Arlecino [84]2 years ago

3 0

3 + .4+.005 is the expanded form

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Please help I am struggling please

Katen [24]

Answer:

x = 8sqrt3

y=8

Step-by-step explanation:

cos60=y/16

y=16cos60

y=16x1/2

y=8

sin60=x/16

x=16sin60

x=(16sqrt3)/2

x=8sqrt3

3 0

2 years ago

What is the equation of the line that passes through the point (-6,-8) and has a slope of 1/3

Arturiano [62]

y + 8 = 1/3 (x+6)

With the given information, we can use the point-slope formula, , to write the equation of the line. Substitute values for the , , and in the formula to do so.

The represents the slope, so substitute in its place. The and represent the x and y values of one point the line intersects, so substitute -6 for and -8 for . This gives the following answer and equation (just make sure to convert the double negatives into positives:

8 0

2 years ago

A draw scaling of a parking lot has a width of four inches. If the scale is 1 inch = 8 feet. What is the actual width of the par

IgorLugansk [536]

Answer: 32 feet

Step-by-step explanation:

Since 1 inch equals 8 feet and the parking lot has a width of 4 inches we multiply 8 times 4 giving you an answer of 32, so the parking lot's width is 32 feet.

4 0

2 years ago

Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two bakeries' specialty pi

DIA [1.3K]

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (49) < μ₁ - μ₂ < (289)

Step-by-step explanation:

The given data are;

Bakery A

$\overline x_1$ = 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

$\overline x_2$ = 1,711 cal

s₂ = 192 cal

n₂ = 10

$\left (\bar{x}_1-\bar{x}_{2} \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2} \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}$

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, $t_c$ = 1.706

$\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}$

$\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469$

$\hat \sigma$ ≈ 178

Therefore, we get;

$\left (1,880-1,711 \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711 \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}$

Which gives;

$169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}$

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

4 0

2 years ago

Help plsssssssssssss

svetlana [45]

Answer:

5. a) 2.140 + 1.725 = 3.865

b) 29.99 + 1.42 = 31.41

c) 6.157 - 1.009 = 5.148

6) 7.4 - 0.0022 = 7.2978 ✅

7.39 - 0.0012 = 7.3888

7. 140minutes / 2 hours and 20 minutes

6 0

1 year ago