45^2 can be found out by 4 x ____ x 100 + ____ = ___

Students heights in inches were obtained as part of a study and a frequency distribution was constructed with the first class ha

polet [3.4K]

The frequency of the second class is 6.

Since the class width is 6 and the lower limit of the first class is 50, this means the first class goes from 50-55. This would put the second class at 56-61. There are 6 data points in this set that would go in the second class.

8 0

2 years ago

What is the height of the building pictured below if its shadow is 100 ft. long?

Evgen [1.6K]

Yeah.. baby, do the SOH CAH TOA with the hula hoop

$\bf 
sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad 
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}

\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
$

so.. hmmm we have in the picture, an angle, and two sides,
actually, is the "adjacent" side, and the "opposite"

so.. which of those ratios give us only
the angle
the adjacent side
the opposite side?

ahhha!, is Ms tangent
so $\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(42^o)=\cfrac{h}{100}$

solve for "h"

5 0

2 years ago

A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice we

kompoz [17]

Answer:

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

$X_{ij}$ = The number which comes up on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

$E(X_{ij})$ = $\frac {1 + 2 + 3 + 4 + 5 + 6}{6}$

= 3.5 ∀ i = 1(1)10 and j = 1(1)20

and,

$E(X^{2}_{ij}$ = $\frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}$

= $\frac {1 + 4 + 9 + 16 + 25 + 36}{6}$

= $\frac {91}{6}$

$\simeq$ 15.166667

so, $Var(X_{ij}$ = $(E(X^{2}_{ij} - {(E(X_{ij})}^{2})$

$\simeq 15.166667 - 3.5^{2}$

= 2.91667

and $\sigma_{X_{ij}}$ = $\sqrt {2.91667}[/tex [tex]\simeq 1.7078261036$

Now we get that,

$Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}$

We get that $Y_{j}'s$ are iid RV's ∀ j = 1(1)20

Let, ${\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}$

So, we get that $E({\overline}{Y}) = E(Y_{j})$

= $E(X_{ij}$ for any i = 1(1)10

= 3.5

and,

$\sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}} = \frac {\sigma_{X_{ij}}}{\sqrt {20}} = \frac {1.7078261036}{\sqrt {20}} [tex]\simeq 0.38$

Hence, the option which best describes the distribution being simulated is given by,

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

6 0

2 years ago

Cooks are needed to prepare for a large party. Each cook can bake either 5 large cakes or 14 small cakes per hour. The kitchen i

PolarNik [594]

Answer:

8

Step-by-step explanation:

In 2 hours

1 cook makes 10 large cakes

24/10 = 2.4

You need 3 cooks to make the large cakes.

In 2 hours

1 cook makes 28 small cakes

130/28 = 4.64

You need 5 cooks to make the small cakes.

3 + 5 = 8

Answer: 8

5 0

2 years ago

Help please! I’ll give brainliest too!

DIA [1.3K]

Answer:

y-intercept: -5

Slope: 4

Equation: y = 4x - 5

Step-by-step explanation:

✔️The y-intercept is the value of y when x = 0. From the table of values, it shows that when x = 0, y = -5. Therefore:

y-intercept (b) = -5

✔️using two pairs from the table, (1, -1) and (2, 3):

$Slope (m) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 -(-1)}{2 - 1} = \frac{4}{1} = 4$

Slope = 4

✔️To write an equation for the function, substitute m = 4, and b = -5 into y = mx + b.

Thus:

y = 4x +(-5)

y = 4x - 5

7 0

2 years ago

45^2 can be found out by 4 x ____ x 100 + ____ = ____

45^2 can be found out by 4 x x 100 + =