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4 years ago

Which statement is true? A)4[3(5+2)]=95-4×3 B)4[3(5+2)] not equal 95-4×3

Mathematics

2 answers:

yanalaym [24]4 years ago

7 0

Answer:

B) 4[3(5+2)] not equal 95-4×3

Step-by-step explanation:

4[3(5+2)]=95-4×3

84=95-4x3

84=83

So since they are not equal B. is the correct answer!

wlad13 [49]4 years ago

7 0

Answer:

Option B.

Step-by-step explanation:

The given expression is

4[3(5 + 2)] = 95 - 4×3

We will solve the left side of the expression first

4[ 3(5 + 2)] = 4[ 3(7) ]

= 4 × 21

= 84

Now we take the right side of the expression

95 - 4×3

= 95 - 12

= 83

Now it's clear that right side of the expression is not equal to the left side of the expression.

Therefore, Option B is the correct option.

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Maru [420]

Answer:

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3 years ago

In Major League Baseball, the American League (AL) allows a designated hitter (DH) to bat in place of the pitcher, but in the Na

hammer [34]

Answer:

Step-by-step explanation:

From the given information:

Let:

$\mu_1$ represent the population mean no. for DH group.

$\mu_2$ represent the population mean no. for no DH group

$n_1$ represent the sample sizes of DH

$n_2$ represent the sample sizes of no DH

Sample size: $n_1 = n_2 = 20$

For both groups, the population standard deviation of runs scored = 2.54

i.e

$\sigma^2_1 = \sigma_2^2 = 2.54$

The null & alternative hypothesis;

$H_o : \mu_1 \le \mu_2 \\ \\ H_a: \mu_1 > \mu_2$

Level of significance:

The test statistics is:

$z = \dfrac{\bar x_1 - \bar x_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma_2^2}{n_2} }}$

where;

$\bar x_1$ = sample mean for DH group

$\implies \dfrac{1}{n_1} \sum \limit ^{n_1}_{i=1} x_1 \\ \\ = \dfrac{1}{20}(0+6+8+2+2+4+7+7+6+5+1+1+5+4+4+5+7+11+10+0) \\ \\ = \dfrac{92}{20}$

= 4.6

$\bar x_2$ = sample mean for no DH group

$\implies \dfrac{1}{n_2} \sum \limit ^{n_1}_{i=1} x_1 \\ \\ = \dfrac{1}{20}(3+6+2+4+0+5+7+6+1+8+12+4+6+3+4+0+5+2+1+4) \\ \\ = \dfrac{83}{20}$

= 4.15

Now:

$z = \dfrac{4.60- 4.15}{\sqrt{\dfrac{2.54^2}{20} + \dfrac{2.54^2}{20} }} \\ \\ = \dfrac{0.45}{\sqrt{0.32258 +0.32258 }} \\ \\ = \dfrac{0.45}{0.803219} \\ \\$

= 0.5602

Since the test is one-tailed by looking at that $H_a:$

The P-value = $P(Z > z)$

$\implies 1 - P(Z \le z) \\ \\ = 1 - P(Z \le 0.5602)$

Using Excel Function " =normdist(z)"; we have":

$= 1- 0.7123$

P-value = 0.2877

Decision rule: To reject $H_o$ , if $p-value < \alpha \ at \ 0.10$

Conclusion: SInce P = 0.2877 which is $> \ \alpha \ at \ 0.10.$

We fail to reject the $H_o$ and conclude that there is insufficient evidence to support the given claim that: $\text{more runs are scored in games for which DH is used.}$

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3 years ago

Question

Rus_ich [418]

9514 1404 393

Answer:

S = 4.5π ≈ 14.14 units

Step-by-step explanation:

The arc length is given by ...

s = rθ . . . . . r = radius; θ = central angle in radians

The angle 135° is ...

135° = 135°(π/180°) = 3π/4 radians

The arc of interest has length ...

s = 6×3π/4 = 9/2π ≈ 14.14 . . . units

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3 years ago

What is 92x43 using standard algorithm <br> Plsss answer

Karo-lina-s [1.5K]

92x43 is 3,956!

I’m not quite show how I am supposed to show the standard algorithm here.

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3 years ago

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53 gal 1 qt + 12 gal 3 qt + 82 gal 1 qt = ?

mario62 [17]

Given:

53 gal 1 qt + 12 gal 3 qt + 82 gal 1 qt

To find:

The sum of the given expression.

Solution:

Let us first add gallons:

53 gal + 12 gal + 82 gal = 147 gal

Now, add quarts.

1 qt + 3 qt + 1 qt = 5 qt

We know that,

4 quarts = 1 gallon

So add 1 gallon to 147 gal.

147 gal + 5 qt = 147 gal + 4 qt + 1 qt

= 147 gal + 1 gal + 1 qt

= 148 gal + 1 qt

= 148 gal 1 qt

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3 years ago

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