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monitta
3 years ago
14

I need to know this what is 4.75 as a fraction can someone help

Mathematics
1 answer:
WINSTONCH [101]3 years ago
8 0

Answer:

4 \frac{3}{4}

Step-by-step explanation:

so we see a wholoe number "4", lets set that aside

we are left with 0.75 right

we can also write this as

75/100

we can simplify this by dividing 25 on both sides so it will becocme:

3/4

*Remember we have to bring the "4" we set aside earlier so final answer is*

4   and  3/4

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3 years ago
Consider the graph of the sixth-degree polynomial function f.
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Answer:

b = 1, c = -1 and d = 4

Step-by-step explanation:

To solve this question the rule of multiplicity of a polynomial is to be followed.

If the multiplicity of a polynomial is even at a point, graph of the polynomial will touch the x-axis.

If the multiplicity of the polynomial is odd, graph will cross the x-axis at that point.

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2 years ago
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The equations of two functions are y =-28x + 8 and y=-29x + 7. Which function is changing more
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y=-29x+7 is changing more quickly.

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

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Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two​ bakeries' specialty pi
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Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (<u>49</u>) < μ₁ - μ₂ < (<u>289)</u>

Step-by-step explanation:

The given data are;

Bakery A

\overline x_1<em> </em>= 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

\overline x_2<em> </em>= 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
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