All categories

3 years ago

Which numbers should be multiplied to obtain 175^2 − 124^22?

1 answer:

3 0

This question seems to me to be an exercise in applying the rule for a difference of squares in order to rewrite the difference of two relatively large numbers as the "simpler" product of other integers.

$175^2-124^2=(175-124)(175+124)=51\cdot299=15,249$

You might be interested in

I got 25 so I wanted to know if I'm correct

kvasek [131]

I feel it should be 30 because x represents the number of lessons, and 30 is multiplied by that.

3 0

3 years ago

Read 2 more answers

I really need help on my engineering homework.

schepotkina [342]

This is more like math homework

7 0

3 years ago

Which is the equation of a line perpendicular to the line y

guajiro [1.7K]

Answer:

is b

Step-by-step explanation:

6 0

3 years ago

An airplane flies 3800 km in 4 hours 45 minutes. What is its average speed in kilometers per hour?

8_murik_8 [283]

Average speed = (distance covered in some time) / (time to cover the distance)

4 hours 45 minutes = 4-3/4 hours = 4.75 hours

Average speed = (3,800 km) / (4.75 hours) = 800 km/hr

3 0

3 years ago

Read 2 more answers

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

3 years ago