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Nadusha1986 [10]
3 years ago
10

Solve if h=5 and y=8 7hy^2+y

Mathematics
1 answer:
dybincka [34]3 years ago
5 0

Step-by-step explanation:

h = 5

y = 8

=> 7hy² + y

=> 7 × 5 × 8² + 8

=> 35 × 64 + 8

=> 2240 + 8

=> 2248

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Byron has his own bakery. He bakes 84 cakes each week. How many can he make in one day?
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Answer: 12 cakes

Step-by-step explanation:

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3 years ago
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A point T on a segment with endpoints D (1, 4) and F (7, 1) partitions the segment in a 3:1 ratio. Find T.
velikii [3]

Step-by-step explanation:

given,

x1= 1 , x2= 7 , m1= 3

y1= 4 , y2= 1 , m2= 1

x=?

y=?

using section formula

x= (m2.x1 + m1.x2)/m1 + m2

x= find yourself

y= (m2.y1 + m1.y2) / m1 + m2

y= find yourself

5 0
2 years ago
An ellipse has a center at the origin, a vertex along the major axis at (0, 17), and a focus at (0, −8). Which equation represen
Marianna [84]
We know that
<span>Since the focus and vertex are above and below each other, rather than side by side, I know that this ellipse must be taller than it is wide.
</span>Then 
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</span>Also, since the focus is 8 <span>units below the center, then </span><span>c = 8
</span>since the vertex is 17<span> units above, then </span><span>a = 17
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</span>b²=17²-8²-----> b²=225
the equation is
y²/a²+x²/b²=1------> y²/289+x²/225=1

the answer is
y²/289+x²/225=1

see the attached figure

7 0
3 years ago
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The bad debt ratio for a financial institution is defined to be the dollar values of loans defaulted divided by the total dollar
Nimfa-mama [501]

Answer:

(a) NULL HYPOTHESIS, H_0 : \mu \leq  3.5%

    ALTERNATE HYPOTHESIS, H_1 : \mu > 3.5%

(b) We conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Step-by-step explanation:

We are given that a random sample of seven Ohio banks is selected.The bad debt ratios for these banks are 7, 4, 6, 7, 5, 4, and 9%.The mean bad debt ratio for all federally insured banks is 3.5%.

We have to test the claim of Federal banking officials that the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

(a) Let, NULL HYPOTHESIS, H_0 : \mu \leq  3.5% {means that the the mean bad debt ratio for Ohio banks is less than or equal to the mean for all federally insured banks}

ALTERNATE HYPOTHESIS, H_1 : \mu > 3.5% {means that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks}

The test statistics that will be used here is One-sample t-test;

                T.S. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where,  \bar X = sample mean debt ratio of Ohio banks = 6%

             s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} } = 1.83%

             n = sample of banks = 7

So, test statistics = \frac{6-3.5}{\frac{1.83}{\sqrt{7} } }  ~ t_6

                             = 3.614

(b) Now, at 1% significance level t table gives critical value of 3.143. Since our test statistics is more than the critical value of t so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Hence, Federal banking officials claim was correct.

7 0
3 years ago
Find the derivative.
Aleksandr [31]

Answer:

Using either method, we obtain:  t^\frac{3}{8}

Step-by-step explanation:

a) By evaluating the integral:

 \frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du

The integral itself can be evaluated by writing the root and exponent of the variable u as:   \sqrt[8]{u^3} =u^{\frac{3}{8}

Then, an antiderivative of this is: \frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}

which evaluated between the limits of integration gives:

\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}

and now the derivative of this expression with respect to "t" is:

\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

g(x)=\int\limits^x_a {f(t)} \, dt

is continuous on [a,b], differentiable on (a,b) and  g'(x)=f(x)

Since this this function u^{\frac{3}{8} is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}

5 0
3 years ago
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