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krek1111 [17]
3 years ago
8

What is the area of a square with side lengths of 7x?

Mathematics
2 answers:
yulyashka [42]3 years ago
6 0

Answer:

49x

Step-by-step explanation:

7x x 7x = 49x

max2010maxim [7]3 years ago
5 0

Answer:

area=l²=(7x)²=49x²is your answer

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Please help with number 12
makkiz [27]

Answer:

Step-by-step explanation:

sin(195º)= -√6+√2/4

cos(195º)=-√6-√2/4

tan(195º)=2-√3

8 0
3 years ago
Read 2 more answers
55.8 is 62% of what number?
zhenek [66]
90

55.8 /? = 62/100
55.8 x 100= 62 x ?
5580=62 x ?
5580/62 = ?
?=90
7 0
3 years ago
Evaluate the expression. 6÷−3<br>A.-2<br>B.-3<br>C.3<br>D.2
olya-2409 [2.1K]
A is the correct awnser good luck
7 0
3 years ago
Dave is comparing the circumferences of several trees in his yard. The oak tree is 0.539 meters in circumference, the ash tree h
stepan [7]

Answer:   c. elm

Step-by-step explanation:

Given : The circumference of the oak tree= 0.539 meters

=0.539\times100=53.9\text{ cm}   [∵ 1 m = 100 cm]             (1)

The circumference of the ash tree= 0.509 yards

=0.509\times3\text{ feet}   [∵ 1 yard = 3 feet]

=0.509\times3\times 30.48\text{ cm}  [∵ 1 foot = 30.48 cm]

=46.54296\approx46.54\text{ cm}        (2)

The circumference of the elm tree = 6281.70 millimeters

=\dfrac{6281.70}{10}=628.17\text{ cm}          (3)

The circumference of the poplar tree = 0.000385 miles

=0.000385 \times5280\text{ feet}   [∵ 1 mile = 5280 feet]

=0.000385 \times5280\times30.48\text{ feet}  [∵ 1 foot = 30.48 cm]

=61.959744\approx61.96\ \text{ cm}              (4)

From (1) , (2) , (3 ) and (4) it is clear that

46.54< 53.9 < 61.96 < 628.17

Hence, the elm tree has the greatest circumference.

5 0
3 years ago
In the Salk vaccine field trial, 400,000 children were part of a randomized controlled double-blind experiment. Just about half
EastWind [94]

Answer:

Step-by-step explanation:

From the given information:

The number of children that were randomly allocated to each vaccination group; n₁ = 200,000

No of polio cases X₁ = 57

Now, in the vaccine group:

the proportion of polio cases is:

\hat p_1 = \dfrac{57}{200000}

=  0.000285

The number of children that were randomly allocated to the placebo group, n₂ = 200,000

No of polio cases X₂ = 142

In the placebo group

the proportion of polio cases is:

\hat p_2 = \dfrac{142}{200000}

Null and alternative hypothesis is computed as follows:

H₀: There is no difference in the proportions of polio cases between both groups.  

H₁: There is a difference in the proportions of polio cases between both groups.

Let assume that the level of significance ∝ = 0.05

The test statistic  can be computed as:

Z = \dfrac{\hat p_1-\hat p_2}{\sqrt{\dfrac{\hat p_1 \hat q_1}{n_1}+ \dfrac{\hat p_2 \hat q_2}{n_2}}}

Z = \dfrac{0.000285-0.000710}{\sqrt{\dfrac{0.000285(1-0.000285)}{200000}+ \dfrac{0.000710(1-0.000710)}{200000}}}

Z = \dfrac{-4.25\times 10^{-4}}{\sqrt{\dfrac{0.000285(0.999715)}{200000}+ \dfrac{0.000710(0.99929)}{200000}}}

Z = - 6.03

P-value =  2P(Z < -6.03)

From the Z - tables

P-value =  2 × 0.0000

= 0.000

We reject the H₀ provided that P-value is very less.

Therefore, we may conclude that there is a difference in the proportions of polio cases between the vaccine group and placebo group not due to chance.

7 0
2 years ago
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