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3 years ago

7

MRS SMITH HAS 12 times as many markers as colored pencil. The total number of markers and colored pencil is 78. How many markers

does Mrs smith have?

2 answers:

Anika [276]3 years ago

3 0

Multiply 78 and 12.
the answer would be 936

larisa [96]3 years ago

3 0

The answer is 72 markers because that 12 times how ,any markers there are. There are 6 markers and 78 - 6 is 72 and 72/6 is 12.

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A 13foot lader i upagenst a tree the bottom of the ladder is 5 feet away form the tree how high is the ladder on the tree

SCORPION-xisa [38]

Think of the 13-ft length of the ladder as the hypotenuse of a right triangle. Represent the horiz. distance from foot of ladder to base of tree by x, or 5 ft.

Represent the vert. dist. from base of tree to top of ladder by y, which is unknown.

Then (13 ft)^2 = (5 ft)^2 + y^2, or

169 ft^2 = 25 ft^2 + y^2. This simplifies to y^2 = 144. Thus y = + 12 feeet.

Note: Please pay attention to your spelling: "lader i up agenst a tree" should be "the top of a 13-ft ladder is placed against a tree."

7 0

2 years ago

Someone lmk what the answer is

Elodia [21]

$f(-7) = -(-7)^2 - 10 \cdot (-7) + 16\\= -49 + 70 + 16\\= 22$

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3 years ago

A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average wei

Arte-miy333 [17]

Answer:

<em>Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom</em>

<em>The null hypothesis is accepted . </em>

<em>Assume the population variances are approximately the same</em>

<u><em>Step-by-step explanation:</em></u>

<u>Explanation</u>:-

Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

<em>The first sample size 'n₁'= 20</em>

<em>mean of the first sample 'x₁⁻'= 17.53 pounds</em>

<em>standard deviation of first sample S₁ = 3.2 pounds</em>

Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

<em>The second sample size n₂ = 24</em>

<em>mean of the second sample "x₂⁻"= 14.89 pounds</em>

<em>standard deviation of second sample S₂ = 2.7 poun</em>ds

<u><em>Null hypothesis</em></u><u>:-</u><u><em>H₀</em></u><em>: The Population Variance are approximately same</em>

<u><em>Alternatively hypothesis</em></u><em>: H₁:The Population Variance are approximately same</em>

<em>Level of significance ∝ =0.05</em>

<em>Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42</em>

<em>Test statistic :-</em>

<em> </em> $t = \frac{x^{-} _{1} - x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }$

<em> where </em> $S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2} }{n_{1} +n_{2} -2}$

$S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}$

<em> substitute values and we get S² = 40.988</em>

<em> </em> $t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24} )}$ <em></em>

<em> </em> t = 1.3622

Calculated value t = 1.3622

Tabulated value 't' = 2.081

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

<u><em>Conclusion</em></u>:-

<em>The null hypothesis is accepted </em>

<em>Assume the population variances are approximately the same.</em>

<em> </em>

<em> </em>

<em> </em>

6 0

2 years ago

How do i solve x+y=2?

Aliun [14]

It’s unsolvable with out more context. What’s the rest of the problem.

8 0

3 years ago

Read 2 more answers

Solve for x ....... thanks !

Len [333]

If we can match teh bases we can solve
because if x=x and xᵃ=xᵇ, we can conclude that a=b

16=2⁴
32=2⁵
rememeber that $(x^m)^n=x^{mn}$

$16^{3x+2}=32^{-2x-7}$
$(2^4)^{3x+2}=(2^5)^{-2x-7}$
$2^{4(3x+2)}=2^{5(-2x-7)}$
2=2 so we conclude that 4(3x+2)=5(-2x-7)

4(3x+2)=5(-2x-7)
expand/distribute
12x+8=-10x-35
add 10x both sides
22x+8=-35
minus 8 both sides
22x=-43
divide both sides by 22
x=-43/22

8 0

2 years ago